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Question 1146372: Find the equation of the radical axis of the two circles x2 + y2 - 4x - 6y - 3 = 0 and x2 + y2 - 12x - 14y + 65 = 0.
Answer by KMST(5398)

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You can put this solution on YOUR website!
The radical axis of two circles can be defined as the locus (set) of the points from where tangents to the two circles would have equal lengths.
It can also be defined as the locus of the points whose power relative to both circles is the same.
Power of a point relative to a circle is d%5E2-r%5E2

, where

, and

.

The definitions are equivalent, because in the right triangles formed by tangent, the radius, and the distance

.
The line segment of length

between the point and the circle center is the hypotenuse and the power of the point is the square of the length of the tangent segment,

UNDERSTANDING THE PROBLEM:
The equations of the circles are written in the general form.
They could be written in a form that allowed us to identify the center and radius of each circle.
For the first circle, that would be %28x-2%29%5E2%2B%28y-3%29%5E2-%283%2B2%5E2%2B3%5E2%29=0

%28x-2%29%5E2%2B%28y-3%29%5E2-%283%2B2%5E2%2B3%5E2%29=0

<-->

, showing the center to be C(2,3) , and the radius to be sqrt%2816%29=4

.
For the second circle, that would be %28x-6%29%5E2%2B%28y-7%29%5E2-%286%5E2%2B7%5E2-65%29=0

<-->

, showing the center to be C'(6,7) , and the radius to be sqrt%2820%29

.
The equations in this form show a sum of squares that is the square of the distance from a point P(x,y) to the center of the circle, minus the square of the radius.
That is the power of point P(x,y) relative to the circle.
For points of a circle it is zero, their power relative to the circle is zero,
but for any point P(x,y) the power of P relative to the first circle can be written as
%28x-2%29%5E2%2B%28y-3%29%5E2-16

, or as its equivalent form, x%5E2%2By%5E2-4x-6y-3

.
So, for all the points P(x,y) whose powers relative to both circles are the same,
x%5E2%2By%5E2-4x-6y-3=x%5E2%2By%5E2-12x-14y%2B65

, and from there we get the equation of the radical axis of the two circles.
cross%28x%5E2%2By%5E2%29-4x-6y-3=cross%28x%5E2%2By%5E2%29-12x-14y%2B65

-->

-->

-->

-->

The circles, with their centers, the line segment connecting those centers, and the radical axis are shown below.

Question 1147397: find the canonical form and determine the nature of conics
8x^2-12xy+17y^2+16x-12y+3=0
Answer by KMST(5398)

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You can put this solution on YOUR website!
A quadratic equation of the form Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

, like

, called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
In the case of 8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0

, with

B%5E2-4AC=%28-12%29%5E2-4%2A8%2A17=144-544=-400%3C0

suggest that the equation represents an ellipse.
If there was no term in

, the axes of symmetry of the ellipse would be parallel to the x- an y-axes.
If there is term in

, one of the axes of symmetry will be at a positive angle alpha

to the positive x-axis such that 0%5Eo%3Calpha%3C90%5Eo

.
Rotating the coordinate axes counterclockwise any angle alpha

such that

,
a point P%28x%2Cy%29

would be called the point P%28u%2Cv%29

with coordinates referencing the new axes,
with u=x%2Acos%28alpha%29%2By%2Asin%28alpha%29

and

v=-x%2Asin%28alpha%29%2By%2Acos%28alpha%29

.
For the reverse conversion, highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29

and

highlight%28y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29

.

The points represented by equation Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

would be represented in the new set of axes
by a new equation in

and

, with new coefficients. Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0

.
Substituting highlight%28u%2Acos%28alpha%29-v%2Asin%28alpha%29%29

for

and

highlight%28u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29

for

into the original equation,
we would get the equation Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0

.
Expressions to calculate the new coefficients can be found, including

we must find a value for alpha

that makes K=0

B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29%2B2%28C-A%29sin%28alpha%29cos%28alpha%29=0

-->

B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29=2%28A-C%29sin%28alpha%29cos%28alpha%29

-->

2sin%28alpha%29cos%28alpha%29%2F%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29

That can be converted into highlight%28B%2F%28A-C%29=sin%282alpha%29%2Fcos%282alpha%29=+tan%282alpha%29%29

by using the trigonometric identities cos%282alpha%29=+cos%5E2%28alpha%29-sin%5E2%28alpha%29

and

sin%282alpha%29=+sin%28alpha%29cos%28alpha%29

From there the values for 2alpha

and its trigonometric functions can be found.
With those values the other coefficients can be calculated as
J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29+

J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29+

L=+A%2Asin%5E2%28alpha%29+-B%2Asin%28alpha%29cos%28alpha%29%2BC%2Acos%5E2%28alpha%29

M=+D%2Acos%28alpha%29%2BE%2Asin%28alpha%29

, and

N=-D%2Asin%28alpha%29%2BE%2Acos%28alpha%29

.
After that, the canonical form for the equation in

and

can be found.

FINDING alpha

AND ITS TRIGONOMETRIC FUNCTIONS:
As a quadratic equation of the form Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

has

system%28A=8%2CB=-12%2CC=17%2CD=16%2CE=12%2CF=3%29

A calculator could provide a good approximation of the value of 2alpha=53.13%5Eo

(rounded) and that value could be used to get approximated values for sin%282alpha%29

, and

.
Alternatively, exact values can be found using trigonometric identities.
Because 0%5Eo%3Calpha%3C90%5Eo

we know that 0%5Eo%3C2alpha%3C180%5Eo

}, but because tan%282alpha%29=4%2F3%3E0

we conclude 0%5Eo%3C2alpha%3C90%5Eo

as well, and al trigonometric functions will be positive.
From sin%282alpha%29%2Fcos%282alpha%29=4%2F3

<-->

sin%282alpha%29=%284%2F3%29cos%282alpha%29

and trigonometric identity sin%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1

we get

%28%284%2F3%29cos%282alpha%29%29%5E2%2Bcos%5E2%282alpha%29=1

-->

%2816%2F9%29cos%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1

-->

-->

-->

-->

cos%282alpha%29=sqrt%289%2F25%29=3%2F5=0.6

and

sin%282alpha%29=%284%2F3%29cos%282alpha%29=%284%2F3%29%283%2F5%29=4%2F5=0.8

.
From

and the trig identities for half angles we can find sin%5E2%28alpha%29

, as

-->

sin%28alpha%29=sqrt%281%2F5%29=sqrt%285%2F25%29=highlight%28sqrt%285%29%2F5%29

-->

cos%28alpha%29=sqrt%284%2F5%29=highlight%282sqrt%285%29%2F5%29

We could also get the product highlight%28sin%28alpha%29cos%28alpha%29=2%2F5%29

from

4%2F5=sin%282alpha%29=2sin%28alpha%29cos%28alpha%29

-->

%281%2F2%29%282sin%28alpha%29cos%28alpha%29%29=%281%2F2%29%284%2F5%29=2%2F5

.

CALCULATING

AND

:
Now we can substitute the highlighted values above into the equations to calculate

, and

8%2A%284%2F5%29%2B%28-12%29%282%2F5%29%2B17%2A%281%2F5%29

, and

.

FINDING THE CANONICAL FORM AND THE NATURE OF THIS CONIC:
For now, substituting the values found for

and

, we can write the equation of the conic in terms of

and

as:

,
and we can and need to transform it into something of the form %28u-h%29%5E2%2Fp%5E2%2B%28v-k%29%5E2%2Fq%5E2=1

which would be equivalent to
%28%28u-h%29%2Fp%29%5E2%2B%28%28v-k%29+%2Fq%29%5E2=1

.
That is the equation of a circle of radius

, stretched in the u and v directions by factors

and

, centered in point (h,k) .
In other words, that represents an ellipse centered in (h,k) .
To get that transformation we need to form the squares %28u-h%29%5E2=u-2uh%2Bh%5E2

and

from

and

respectively .
5u%5E2%2B20v%5E2%2BMu%2BNv%2B3=0

-->

-->

M%2F10=4sqrt%285%29%2F10=highlight%280.4sqrt%285%29%29=0.89443

,
and

N%2F40=-8sqrt%285%29%2F40=highlight%28-0.2sqrt%285%29%29=-0.447215

%28M%2F10%29%5E2=%280.4sqrt%285%29%29%5E2=0.16%2A5=0.8

%28N%2F40%29%5E2=%28-0.2sqrt%285%29%29%5E2=0.04%2A5=0.2

Then, the right hand side term is -3%2B5%28M%2F10%29%5E2%2B20%28N%2F40%29%5E2

-3%2B5%2A%280.4sqrt%285%29%29%5E2%2B20%2A%280.5sqrt%285%29%29%5E2

-3%2B5%2A0.16%2A5%2B20%2A0.04%2A5=-3%2B4%2B4=5

, and the equation is
highlight%285%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v%2B%28-0.2sqrt%282%29%29%29%5E2=5%29

From there, we continue:
5%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v-0.2sqrt%285%29%29%5E2=5

-->

%28u%2B0.4sqrt%285%29%29%5E2%2B4%28v-0.2sqrt%282%29%29%5E2=1

-->

%28u-%28-0.4sqrt%285%29%29%29%5E2%2F1%5E2

The center of the ellipse is at %22%28%22

.
The numbers squared in the denominators are the semi-major and semi-minor axes of the ellipse.
The greater one,

, is the semi-major axis, and the other I the semi-minor axis
The extreme values for

happen when v-0.2sqrt%285%29=0

, and then
%28u%2B0.4sqrt%285%29%29%5E2=1

-->

-->

.
The y-coordinate of the center and the extremes show the vertices are at a distance of

from the center .
Using the equations highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29

and

, we can find the coordinates of the center and vertices of the ellipse.
For the center, x=%28-0.4sqrt%285%29%29%2A%282sqrt%285%29%2F5%29-0.2sqrt%285%29%2A%28sqrt%285%29%2F5%29

x=%28-0.4sqrt%285%29%29%2A%282sqrt%285%29%2F5%29-0.2sqrt%285%29%2A%28sqrt%285%29%2F5%29

-0.8%2A5%2F5-0.04sqrt%2825%29=-0.8-0.04%2A5=-0.8-0.2=-1

y=-0.4sqrt%285%29%2A%28sqrt%285%29%2F5%29%2B0.2sqrt%285%29%2A%282sqrt%285%29%2F5%29

-0.4%2A5%2F5%2B0.2%2A2%2A5%2F5=-0.4%2B0.4=0

Question 1150043: Write an equation for the hyperbola that has eccentricity 2, center at (0,0), and vertex at (0,8)
An equation for the hyperbola is...?
Answer by KMST(5398)

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You can put this solution on YOUR website!
FACTS ABOUT HYPERBOLAS:
The equation for a hyperbola centered at (h,k) can be written as
%28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=%22+%22%2B-+1

%28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=%22+%22%2B-+1

for some pair (p,q) of positive numbers.
If a given equation can be written as above, you would have the values of h, k, p, and q and the sign for the %22+%22%2B-+1

part.
For %28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=-1

<-->

%28y-k%29%5E2%2Fq%5E2-%28x-h%29%5E2%2Fp%5E2=1

<-->

%28y-k%29%5E2%2Fq%5E2=%28x-h%29%5E2%2Fp%5E2%2B1

, you see that it must be %28y-k%29%5E2%2Fq%5E2%3E=1

-->

-->

system%28y-k%3E=q%2C%22or%22%2Cy-k%3C=-q%29

-->

system%28y%3E=k%2Bq%2C%22or%22%2Cy%3C=k-q%29

,
so the graph will have an upper branch and a lower branch, like this:

On the other hand, for %28x-h%29%5E2%2Fp%5E2-%28y-k%29%5E2%2Fq%5E2=1

<-->

%28x-h%29%5E2%2Fp%5E2=%28y-k%29%5E2%2Fq%5E2%2B1

, so it must be %28x-h%29%5E2%2Fp%5E2%3E=1

<-->

-->

system%28x-h%3E=p%2C%22or%22%2Cx-h%3C=-p%29

-->

system%28x%3E=h%2Bp%2C%22or%22%2Cy%3C=h-p%29

,
so there is a left branch and a right branch to the graph, like this:
:

.
In either case, the red and green lines are the asymptotes, with slopes q%2Fp

and

.
Textbooks would write the equation of a hyperbola as
either %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1

for a hyperbola with upper and lower branches with vertices at y=k+%2B-+a

,
or

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

for a hyperbola with upper and lower branches with vertices at x=h+%2B-+a

.
When you write the equation you have number values instead of letters,
but your teacher may expect you to use the same letters they use.

COMPARING THOSE FACTS WITH YOUR QUESTION:
The hyperbola in your question has its center at (0,0), and vertex at (0,8), directly above its center by 8 units.
It is just like the hyperbola with equation %28y-k%29%5E2%2Fq%5E2-%28x-h%29%5E2%2Fp%5E2=1

shown in the top drawing above.
The center for the hyperbola in the equation above was (h,k), so for the hyperbola in your question, (h,k) is (0,0), with h=k=0

.
The distance between the center and the vertex above was labeled as

in the drawing, so for the hyperbola in your question, q=8

.
All you need to write your equation is the value of

.

ABOUT ECCENTRICITY:
So, what do we know about eccentricity?
It is defined for ellipses and hyperbolas as a ratio of distances from the center, distance to a focus divided by distance to a vertex to the focus.
and for both curves the drawings have right triangles with sides labeled

, and

, or

, and

,where

is the hypotenuse and the distance from center to focus.
According to the Pythagorean theorem, for those triangles c%5E2=p%5E2%2Bq%5E2

In the drawing above, the triangle side labeled c is not the line segment from the center to a focus, but I draw an arc of circle to show that the lengths are the same.

FINDING

AND

With the equation written as y%5E2%2F8%5E2-x%5E2%2Fb%5E2=1

,
the eccentricity would be e=c%2Fa

, so

-->

-->

Substituting the values found got

and

becomes

-->

b%5E2=16%5E2-8%5E2=256-64=highlight%2892%29

and the equation for the hyperbola in the question,
y%5E2%2F8%5E2-x%5E2%2Fb%5E2=1

turns into highlight%28y%5E2%2F64-x%5E2%2F192=1%29

The graph is

. The asymptotes' slopes are red%28a%2Fb=8%2F8sqrt%283%29=1%2Fsqrt%283%29=sqrt%283%29%2F3%29

and

green%28-a%2Fb=-1%2Fsqrt%283%29=-sqrt%283%29%2F3%29

Question 1152643: Find the value/s of U so that the graph of the equation of Ux2 + y^2 − 2Ux = 0 is a hyperbola.
Afterwards, identify the distance/s between the foci.
Answer by KMST(5398)

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You can put this solution on YOUR website!
I believe the equation was meant to be Ux%5E2+%2B+y%5E2+-+2Ux+=+0

.

FACTS ABOUT HYPERBOLAS:
The equation for a hyperbola centered at (h,k) can be written as
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=%22+%22%2B-+1

for some pair (a,b) of positive numbers.
For %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=-1

<-->

%28y-k%29%5E2%2Fb%5E2-%28x-h%29%5E2%2Fa%5E2=1

<-->

%28y-k%29%5E2%2Fb%5E2=%28x-h%29%5E2%2Fa%5E2%2B1

, you see that it must be %28y-k%29%5E2%2Fb%5E2%3E=1

-->

, so the graph will have an upper branch and a lower branch, like this:

The red and green lines are the asymptotes, with slopes b%2Fa

and

.
In the other hand, for %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

<-->

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2%2B1

, so it must be %28x-h%29%5E2%2Fa%5E2%3E=1

<-->

, so thee is a left branch and a tight branch to the graph, that looks like this:

SOLVING THE PROBLEM:
Adding

to both sides of Ux%5E2+%2B+y%5E2+-+2Ux+=+0

, we get

-->

-->

-->

highlight%28%28x-1%29%5E2%2By%5E2%2FU=1%29

That equation represents a hyperbola for any negative value of

.
If allowed to choose a value for

to calculate distances, I would choose U=-1

.
That makes the equation %28x-1%29%5E2-y%5E2=1

, matching the general equation %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

with

.
The equation represents a hyperbola with center (1,0),
vertices V%280%2C0%29

and

, asymptotes y=%22+%22%2B-+%28x-1%29

, focal distance c=sqrt%28a%5E2%2Bb%5E2%29=sqrt%282%29

, and foci

and

.
The distance between the foci is 2sqrt%282%29

If a generic

is expected, we can say U=-b%5E2

for any

Then we have %28x-1%29%5E2-y%5E2%2Fb%5E2=1

.
That equation represents a hyperbola with center (1,0),
vertices V%280%2C0%29

and

, asymptotes y=%22+%22%2B-+b%28x-1%29

, focal distance c=sqrt%281%2Bb%5E2%29

, and foci at %22%28%22

.
The distance between the foci is 2sqrt%281%2Bb%5E2%29

Question 1158804: 9x^2-24xy+16y^2-20x-15y-50=0
Use axis rotation formulas for x and y to transform the quadratic equation to an equation in (u,v) coordinates with no cross product term. Identify the vertex or vertices in (x,y) coordinates
thank you so so so much
Answer by KMST(5398)

(Show Source):

You can put this solution on YOUR website!
A quadratic equation of the form Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

, like

, could represent a circle, ellipse, hyperbola, parabola.
In special cases it could represent a point, a line, a pair of lines, or no point that could exist, depending on the values of the coefficients.

ROTATING AXES:
Axis rotation equations:
From coordinates (x, y), to get the new coordinates (u, v) that would result for a point (x,y) after rotating the coordinate axes counterclockwise an angle +alpha+

we can apply
u=x%2Acos%28alpha%29%2By%2Asin%28alpha%29

From (u, v), rotating an angle -alpha

(the reverse change) we get (x,y) using
x=u%2Acos%28-alpha%29-v%2Asin%28-alpha%29

-->

y=-u+%2Asin%28-alpha%29%2Bv%2Acos%28-alpha%29%29

-->

Substituting u%2Acos%28alpha%29-v%2Asin%28alpha%29

for

and

+u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29

for

,
with a lot of busy algebra work, we could get a new equation in

and

with new expressions for the new coefficients.
That is grueling, mistake-inducing work (and torture to type and edit).
We would have expressions involving alpha

for all the new coefficients of terms with

and/or

, including the coefficient of the term in

.
That coefficient must be made equal to zero.
From the equation making the coefficient of

equal to zero, a value for alpha

needs to be found to be used to calculate all other coefficients.
Maybe that is the initial work that was expected for this problem.

IDENTIFYING VERTEX (OR VERTICES):
The new equation on

and

should be identified as a parabola (one vertex), a hyperbola (two vertices), an ellipse (two vertices and two co-vertices), or something else.
Then, the (u,v) coordinates for any vertices should be found, and converted into (x,y) coordinates.

POSSIBLE SHORTCUTS:
Rotating the coordinate axes counterclockwise an angle alpha

, from an equation
Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

we get a new equation in

and

with new coefficients:
Jx%5E2%2BKxy%2BLy%5E2%2BMx%2BNy%2BF=0

.
The new coefficients can be found to be:
J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29+

, and

.
To eliminate the term in

we must find a value for alpha

that makes

using the trigonometric identities

Then, it becomes B%2Acos%282alpha%29=%28A-C%29%2Asin%282alpha%29

-->

+B%2F%28A-C%29=sin%282alpha%29%2F+cos%282alpha%29

-->

system%28B%2F%28A-C%29=X%2C+sin%282alpha%29=cos%282alpha%29%28B%2F%28A-C%29%29%29

. B/(A-C)=tan(2alpha), sin(2alpha)=cos(2alpha)(B/(A-C)
Avoiding mistakes, with careful algebra work, I might have been able to find the expressions outlined above.
I just copied them from an old Analytical Geometry textbook.
The value B%5E2-4AC

, called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
I got that from the book too.
In the case of 9x%5E2-24xy%2B16y%5E2-20x-15y-50=0

B%5E2-4AC=24%5E2-4%2A9%2A16=3%5E2%2A8%5E2-9%2A8%5E2=576-576=0

says the equation represents a parabola.

A SHORTCUT THAT SHOULD BE ALLOWED:
Using

, we can translate into (u, v) coordinates the terms of degree 2 (the first 3 terms):
9x%5E2=9%28u%2Acos%28alpha%29-v%2Asin%28alpha%29%29%5E2

16y%5E2=16%28u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29%5E2

When “collecting like terms”, the term in

will be

-18uv%2Asin%28alpha%29cos%28alpha%29+%2B32uv%2Asin%28alpha%29cos%28alpha%29

.
Making the coefficient of

equal to zero, we get the equation
-18sin%28alpha%29cos%28alpha%29+%2B32sin%28alpha%29cos%28alpha%29=0

We can simplify that equation using the trigonometric identities
cos%282alpha%29=cos%5E2%28alpha%29-sin%5E2%28alpha%29

and

sin%282alpha%29=2sin%28alpha%29cos%28alpha%29

.
We get

-->

system%28sin%282alpha%29=%2824%2F7%29cos%282alpha%29%2C+and%2C+tan%282alpha%29=24%2F7%29

From that we would not get an exact value for 2alpha

, but we know that alpha is in.
A calculator would give you an approximation with enough digits to figure out that cos%28alpha%29=0.8

and

.
We can get exact values of the trigonometric functions and their squares, from sin%282alpha%29=%2824%2F7%29cos%282alpha%29

-->

%28%2824%2F7%29cos%282alpha%29%29%5E2%2Bcos%5E2%282alpha%29=1

-->

%28576%2F49%29cos%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1

-->

-->

-->

-->

and

highlight%28cos%282alpha%29=7%2F25=cos%5E2%28alpha%29-sin%5E2%28alpha%29%29

Then,

-->

sin%282alpha%29=%2824%2F7%29%287%2F25%29=24%2F25

, so

highlight%28sin%282alpha%29=24%2F25=2sin%28alpha%29cos%28alpha%29%29

The sine and cosine of alpha

can also be calculated from cos%282alpha%29

using the trigonometric identities for half angles as

-->

and

-->

Now we can go back to expressions for the terms in u%5E2

and

, and substitute for the trigonometric functions involved the values we found and highlighted before.
As found before they come from the original terms 9x%5E2

, and

expressed as a function of

and

highlight%28sin%282alpha%29=24%2F25-sin%28alpha%29cos%28alpha%29%29

and

,
and

Collecting like terms, we find the term in u%5E2

to be

red%289%284%2F5%29%5E2%28alpha%29-24%284%2F5%29%283%2F5%29%2B16%2A%283%2F5%29%5E2%29u%5E2

red%289%2A16%2F25-12%2812%2F25%29%2B16%2A%289%2F25%29%29u%5E2

,
and the tern in v%5E2

to be

blue%289%283%2F5%29%5E2%2B24%283%2F5%29%284%2F5%29%2B16%284%2F5%29%5E2%29v%5E2

blue%289%2A9%2F25%2B24%2A12%2F25%2B16%2816%2F25%29%29v%5E2

The linear (degree 1) terms in

and

come from

-20x=-20%28u%2Acos%28alpha%29-v%2Asin%28alpha%29%29

-20%28%284%2F5%29u-%283%2F5%29v%29=-16u%2B12v

, and

-15y=-15%28u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29

-15%28%283%2F5%29u%2B%284%2F5v%29%29=-9u-12v

The equation on

and

red%280%29%2Au%5E2%2B0%2Auv%2Bblue%2825%29v%5E2-25u%2B0v-50=0

, which simplifies to v%5E2-u-2=0

, representing a parabola with axis v=0

, and vertex V%28-2%2C0%29

all in u,v coordinates.
The x,y coordinates of the vertex are
x=u%2Acos%28alpha%29-v%2Asin%28alpha%29=-2%2A%284%2F5%29-0%2A%283%2F5%29=-8%2F5=-1.6

y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29=-2%2A%283%2F5%29%2B0%2A%284%2F5%29=6%2F5=-1.2

Question 976252: a one-way road passes under an overpass in the shape of half and ellipse, 30ft high at the center and 20ft wide. Assuming a truck is 12ft wide, what is the tallest truck that can pass under the overpass?
Answer by ikleyn(53937)

(Show Source):

You can put this solution on YOUR website!
.
a one-way road passes under an overpass in the shape of half and ellipse, 30ft high at the center and 20ft wide.
Assuming a truck is 12ft wide, what is the tallest truck that can pass under the overpass?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @lwsshak3 are incorrect, leading to wrong answer.
        I came to provide a correct solution.

Ellipse has a vertical major axis
Its basic form of equation:
x^2/b^2+y^2/a^2=1
For given problem:
a=30
b=10
equation: x^2/10^2+y^2/30^2=1
=x^2/100+y^2/900=1
coordinates of point at which a 12 ft wide truck would reach the overpass=(6,y)
plug in x-coordinate and solve for y
y^2/900=1-x^2/100=1-36/100=64/100         <<<---=== this is the point where @lwsshak3 made his mistake
y^2=9*64=576
y=√576 = 24

what is the tallest truck that can pass? < 24 ft

More accurately, the conception "tallest truck" does not work in this problem,
so the question is posed incorrectly in the post.

A correct question should ask about the set of possible heights.
Then the answer " h < 24 ft " is correct and accurate.

Solved correctly.

Question 46730: The graph of xsquare unit +y square unit =9 is translated 4 units to hte right and 5 units up. Find the equation of the new circle.
Answer by ikleyn(53937)

(Show Source):

You can put this solution on YOUR website!
.
The graph of x square unit +y square unit =9 is translated 4 units to hte right and 5 units up.
Find the equation of the new circle.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The answer in the post by @consc198 is incorrect and irrelevant.
See the correct solution in my post below.

The original graph is for equation   +  = 9.


It is the graph of the circle of the radius 3 units centered at the beginning of the coordinate system.


After translation described in the post, it becomes the circle of the radius 3 units centered at the point (4,5).


The standard equation of such translated circle is

     +  = 9.    ANSWER

Solved correctly.

As usual, the "solution" b y @consc198 has nothing in common with the truth
and is inacceptable to teach students.

Question 41391: y=2sin(1/4)x
Found 2 solutions by ikleyn, ikleyn:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
This question is from textbook
y=2sin(1/4)x
~~~~~~~~~~~~~~~~~~~~~~~~~

Dear visitor, if you read your post, you should get understanding that THERE IS NO question in your post.

There is some formula, but the question itself is lost/missed.

Nevertheless, @consc198 in his post gives the answer " 0 " for non-existing question.

May be, he is able to read behind the written text or between the lines ?

In any case, it looks more like gibberish than reasonable communication.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
This question is from textbook
y=2sin(1/4)x
~~~~~~~~~~~~~~~~~~~~~~~~~

Dear visitor, if you read your post, you should get understanding that THERE IS NO question in your post.

There is some formula, but the question itself is lost/missed.

Nevertheless, @consc198 in his post gives the answer " 0 " for non-existing question.

May be, he is able to read behind the written text or between the lines ?

In any case, it looks more like gibberish than reasonable communication.

Question 47802: find all real solutions: 2squaredx-24(2squared)=256
i know this doesnt make much sense. i had to put squared in as words because i cant get the symobol you know the small 2
thanks for all fo your help
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.

What is written in your post ?

Hello, we just live in the 21-th century.

If you write Math and use a computer, it is assumed that you are able
to write Math on a computer keyboard clearly and accurately in a readable way.

As you write in your post, it is a blatant gibberish.
People do not write this way just three hundred years.

Question 1164281: A flashlight is shaped like a paraboloid, so that if its light bulb is placed at the focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the axis. If the paraboloid has a depth of 1.8 in and the diameter on its surface is 6 in, how far should the light source be placed from the vertex? (Round off your answer to 2 decimal places and don't include the unit)

Answer by KMST(5398)   (Show Source):
You can put this solution on YOUR website!
I already posted a solution of the same problem that was also posted as Question 1164605.
I will add context to relate concepts algebra, geometry, precalculus, and physics that are taught separately, and with little relation to history or to the real world.

A paraboloid is a bowl-shaped surface, like the surface swept by a parabola as it rotates around its axis of symmetry.
A cross section "cutting" through a plane containing the axis is a parabola.
If we place the x- and y-axes with the axis of the paraboloid along the y-axis and the origin at the apex of the paraboloid,
the cross section would be a parabola that we can represent by the familiar equation , and a graph like the one below:

THE DEFINITION OF A PARABOLA is the set of points in a plane that are at equal distance from a point called the ,
and a line called the , as shown in the figure above.
That definition is about 1700 years old, but wise ancient Greeks knew parabolas some 2300 years ago, based on their knowledge of geometry and ratios.
Other civilizations had similarly wise men, developing mathematical concepts independently.
The idea of the coordinate plane came later, and it took longer for the standardization of the ways of representing and communicating math accepted across cultures that we use now.
When represented on an x-y coordinate plane, the directrix of a parabola can be any line, but to make equations easier, I chose to put the vertex of the parabola at the origin and to make the directrix parallel to the x-axis.
For that parabola, those choices give us an easy equation : and a focal distance distance ,
the focus will be at point , and the directrix will be the line , at the same distance from vertex .
Considering the point of the parabola with , , with , we see that its distance to directrix is .
That must be the same distance to focus , so .
The point , being a point of the parabola with equation ,
it must satisfy that equation, so
-->-->-->-->--><-->
The relations highlighted are not formulas to memorize. They will be remembered as long as they are being used often, but can be easily discovered/rediscovered as needed.

BACK TO THE PROBLEM:
As we get further away from the apex/vertex, a paraboloid (and a parabola) go on widening forever, but the reflecting surface of the flashlight is just a finite piece of a paraboloid with a depth of 1.8 inches and a diameter of 6 inches at the open end.
The corresponding graph would follow the parabola, but end at the points (-3,1.8) and (3,1.8).
The endpoints with and would satisfy , so
, and
From there, if we like to remember formulas, we could use the one that says that
the focal distance of a parabola with the equation
is to calculate
that is the the distance from the focus of the parabola to its vertex,
from the focus of the paraboloid to its apex,
and the distance we need between the light source to the center of the bottom (apex) of the paraboloid-shaped reflective "bowl".

PHYSICS OF REFLECTION ON PARABOLIC SURFACES:
Reflection is simple. Smooth surfaces can be planar, or have a tangent plane, and there is a normal direction perpendicular to that plane.
Any ray aimed at a reflecting surface at an angle with respect to the normal direction, is reflected with the same angle to the opposite side.
The parabola and parabolic reflecting surfaces were known to wise ancient Greeks about 2300 years ago.
The parabola, reflection, and the properties of paraboloid reflecting surfaces were known to wise ancient Greeks about 2300 years ago.
They knew that they could use a paraboloid reflecting surface to reorganize light coming in all directions from a point source located at the focus of the paraboloid into a beam of rays all going in the direction of the paraboloid axis of symmetry.
They also knew that they could use a paraboloid reflecting surface to concentrate a beam of rays parallel to the paraboloid axis of symmetry into a point at the focus of the paraboloid.
How does that work?
Ray reflects as ray ,or
ray reflects as ray .

Question 1164605: A flashlight shaped like paraboloid, so that if the bulb is focused, the light rays from the bulb is faced at the focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the axis. If the paraboloid has a depth of 1.8 and the diameter is 6 in, how far should the light source be placed from the vertex?

Found 2 solutions by greenestamps, KMST:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

The response from the other tutor is fine. In that response, she uses y = ax^2 for the equation of the parabola with the given dimensions to find that a = 0.2; then she uses the formula f = 1/(4a) to find that the distance from the vertex to the focus is 1.25 inches.

I and many students are familiar with a slightly different convention which of course leads to the same result.

The basic standard formula for the equation of a parabola that I am familiar with is , where a is the distance from the vertex to the focus. Using that to find the distance from the vertex to the focus directly...

ANSWER: 1.25 inches

Answer by KMST(5398)   (Show Source):
You can put this solution on YOUR website!
A paraboloid is a bowl-shaped surface, like the surface swept by a parabola as it rotates around its axis of symmetry.
A cross section "cutting" through a plane containing the axis is a parabola.
If we places the axes with the axis of the paraboloid along the y-axis and the origin at the apex of the paraboloid,
the cross section would be a parabola that we can represent by the familiar equation , and a graph like the one below:

A paraboloid (and a parabola) go on widening forever, but the reflecting surface of the flashlight is just a finite piece of a paraboloid with a depth of 1.8 inches and a diameter of 6 inches at the open end.
The corresponding graph would follow the parabola, but end at the points (-3,1.8) and (3,1.8).
The endpoints with and would satisfy , so
, and
From there, if we like to remember formulas, we could use the one that says that
the focal distance of a parabola with the equation
is to calculate
that is the the distance from the focus of the parabola to its vertex,
from the focus of the paraboloid to its apex,
and the distance we need between the light source to the center of the bottom (apex) of the paraboloid-shaped reflective "bowl".

Question 1164537: 1)Find a new representation of the given equation after rotating through the given angle.
11x^2-50√3xy-39y^2+576=0 60degrees
Answer by KMST(5398)   (Show Source):
You can put this solution on YOUR website!
A quadratic equation of the form could represent a circle, ellipse, hyperbola, parabola. There are special cases where I could represent a point, a line, a pair of lines or no point that could exist, depending on the values of the coefficients.

In the case of we have
, , , and .
The value , called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
suggest it's a hyperbola.
Symmetry and other considerations help visualize the curve represented too.
We can see that if a point (x,y) satisfies that equation, the point (-x, -y) will also satisfy that equation. That tells us that the set of points satisfying that equation is symmetrical with respect to the origin.
When , the equation turns into , and there is here is no solution for , so we know that the curve does not touch or cross the y-axis. That, for an equation with and makes me suspect an hyperbola.

Rotating such a curve (counterclockwise) by an angle we would get another equation with the coefficients A', B', C', and F' replacing A, B, C, and F.
The formulas to find the new coefficients are:
A'=
B'=
C'=
F'=F
, so and
A'=
B'=
C'=
F'=576
The rotated curve equation is , which can be rewritten as
or
That represents a hyperbola centered at the origin, with vertices at (0,-4) and (0,4) and asymptotes
Her is what the two branches of the hyperbola and its asymptotes look like:

Question 443624: Write the slope-intercept equation for the line that passes through (-7, 6) and is perpendicular to -7x + 9y = -2 Please show all of your work.
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Write the slope-intercept equation for the line that passes through (-7, 6) and is perpendicular to -7x + 9y = -2
Please show all of your work.
~~~~~~~~~~~~~~~~~~~~~~~~

        Strictly saying, the solution and the answer in the post by @mananth are incorrect.

        Indeed, if I substitute coordinates (-7,6) into his equation y = -1.29x-3,
        I will get y = -1.29*(-7) + 6 = 6.03, but not precisely 6,
        which means that the point (-7,6) does not lie on the line.

        In this my post, I will present a correct precise solution and the correct answer, and then will explain,
        WHY the solution by @mananth is wrong.

Any line perpendicular to the given line -7x + 9y = -2 has an equation of the form 7y + 9x = c (1) where 'c' is some real constant. So, we only need to find the value of 'c'. To do it, we substitute the coordinates of the given point (-7,6) into equation (1) 7*6 + 9*(-7) = 42 - 63 = -21. So, the sough equation is 7y + 9x = -21. It is NOT that equation, which is presented as the answer in the post by @mananth.

So, I solved the problem correctly, and my answer is different from that by @mananth.

The error made by @mananth, is that he replaced the precise value m1 = of the slope
of the original line by the decimal value 0.78.

This value, 0.78, is only approximation, but not the precise rational value.

So, the error by @mananth is that he uses his computer code in the form, which IS NOT APPLICABLE
and IS NOT ADEQUATE to the problem.

Question 729528: what is the center, verticies, foci and eccentricity of 3x^2 +2y+18x-2y-8=0
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
what is the center, verticies, foci and eccentricity of 3x^2 +2y+18x-2y-8=0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As the post is presented, it is incorrect and is self-contradictory.

For example, it requests to determine the center, but the shape
which corresponds to the given equation has no center.
It also has no eccentricity.

Such "writers" who produce fake content and distribute it on the Internet
should be ruthlessly expelled from the profession.

Question 1165107: Find the equation of the parabola given vertex at (4, 3), axis of symmetry 𝑦 = 3, length of the latus rectum is 4, and 4𝑝 > 0
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
# Given information
h = 4 # x-coordinate of the vertex
k = 3 # y-coordinate of the vertex
latus_rectum_length = 4
condition = "4p > 0"
# The axis of symmetry is y = k (y = 3), so the parabola is horizontal,
# with the standard form: (y - k)^2 = 4p(x - h)
# The length of the latus rectum is |4p|.
# |4p| = 4.
# Since 4p > 0, we must have 4p = 4.
four_p = latus_rectum_length
# The equation is (y - k)^2 = 4p(x - h)
# Substitute the values
equation_lhs = f"(y - {k})^2"
equation_rhs = f"{four_p}(x - {h})"
print(f"Vertex (h, k): ({h}, {k})")
print(f"Axis of symmetry: y = {k}. (Horizontal parabola, form (y - k)^2 = 4p(x - h))")
print(f"Length of latus rectum |4p|: {latus_rectum_length}")
print(f"Condition: {condition}")
print(f"Since |4p| = {latus_rectum_length} and 4p > 0, we have 4p = {four_p}")
print(f"The equation of the parabola is: {equation_lhs} = {equation_rhs}")
```
```text?code_stdout&code_event_index=2
Vertex (h, k): (4, 3)
Axis of symmetry: y = 3. (Horizontal parabola, form (y - k)^2 = 4p(x - h))
Length of latus rectum |4p|: 4
Condition: 4p > 0
Since |4p| = 4 and 4p > 0, we have 4p = 4
The equation of the parabola is: (y - 3)^2 = 4(x - 4)
```
The equation of the parabola can be found using the given information:
1. **Identify the Standard Form:**
* The vertex is at $(h, k) = (4, 3)$.
* The axis of symmetry is the horizontal line $y = 3$.
* Since the axis of symmetry is horizontal ($y=k$), the parabola opens horizontally, and its standard form is:
$$(y - k)^2 = 4p(x - h)$$
2. **Substitute the Vertex Coordinates:**
* Substituting $h=4$ and $k=3$ into the standard form gives:
$$(y - 3)^2 = 4p(x - 4)$$
3. **Determine the Value of $4p$ (Latus Rectum):**
* The length of the latus rectum is $|4p|$.
* We are given that the length of the latus rectum is $4$, so $|4p| = 4$.
* This implies $4p = 4$ or $4p = -4$.
* The condition $4p > 0$ tells us that we must choose the positive value:
$$4p = 4$$
4. **Final Equation:**
* Substitute $4p = 4$ back into the equation from step 2:
$$(y - 3)^2 = 4(x - 4)$$
The equation of the parabola is $\mathbf{(y - 3)^2 = 4(x - 4)}$.

Question 1165709: arch in an amusement park has a parabolic shape. it has a height of 74 feet and the base width of 38 feet. what is the height of the Arch 8 feet from the center?
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
# Given parameters
H = 74 # Maximum height (feet)
W = 38 # Base width (feet)
x = 8 # Horizontal distance from the center (feet)
# The parabola is defined by y = a*x^2 + H.
# The base endpoints are at (W/2, 0) and (-W/2, 0).
# W/2 = 38 / 2 = 19
# We use the point (19, 0) to find 'a'.
# 0 = a * (19)^2 + 74
# a = -74 / 19^2 = -74 / 361
# The equation of the parabola is: y = (-74/361) * x^2 + 74
# Calculate the height (y) at x = 8
height = (-74 / 361) * (x**2) + H
print(f"Height at 8 feet from the center: {height} feet")
```
```text?code_stdout&code_event_index=2
Height at 8 feet from the center: 60.880886426592795 feet
```
The height of the arch 8 feet from the center is approximately **60.88 feet**.
Here is the step-by-step calculation:
### 1\. Establish the Equation of the Parabola
We model the parabolic arch using a standard coordinate system where the vertex (maximum height) is on the $y$-axis and the base is on the $x$-axis.
The general equation for a parabola opening downward with a vertex at $(0, h)$ is:
$$y = ax^2 + h$$
* **Height ($h$):** $74$ feet.
* **Base Width:** $38$ feet, meaning the arch touches the ground (where $y=0$) at $x = \pm \frac{38}{2} = \pm 19$.
Using the point $(19, 0)$ and the vertex $h=74$:
$$0 = a(19)^2 + 74$$
$$0 = 361a + 74$$
$$a = -\frac{74}{361}$$
The equation for the arch is:
$$y = -\frac{74}{361}x^2 + 74$$
### 2\. Calculate the Height
To find the height $y$ at a horizontal distance of $x=8$ feet from the center:
$$y = -\frac{74}{361}(8)^2 + 74$$
$$y = -\frac{74}{361}(64) + 74$$
$$y = 74 - \frac{4736}{361}$$
$$y \approx 74 - 13.11911$$
$$y \approx 60.88088 \text{ feet}$$

Question 1165937: solve for the equation of the hyperbola in general form with directrices x=-5 and x=7, and one focus at (13,5).
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
To find the equation of the hyperbola in general form, we first need to determine the **center**, the distance from the center to the **focus ($c$)**, and the distance from the center to the **directrix ($a/e$)**.
Since the directrices are vertical lines ($x=-5$ and $x=7$), the hyperbola is **horizontal**, meaning its transverse axis is parallel to the $x$-axis.
## 1. Determine Key Parameters
### A. Center ($h, k$)
The center ($h, k$) lies exactly halfway between the two directrices. Since the directrices are defined by $x=-5$ and $x=7$, the $x$-coordinate of the center ($h$) is their average. The focus given is $(13, 5)$, so the $y$-coordinate of the center ($k$) must be 5.
$$h = \frac{-5 + 7}{2} = \frac{2}{2} = 1$$
$$\text{Center} (h, k) = \mathbf{(1, 5)}$$
### B. Distance to Focus ($c$)
The focus is at $(13, 5)$ and the center is at $(1, 5)$. The distance $c$ is the distance between them:
$$c = 13 - 1 = \mathbf{12}$$
### C. Distance between Directrices
The distance between the directrices is $7 - (-5) = 12$. The general distance between the two directrices of a hyperbola is given by $2a/e$.
$$2\frac{a}{e} = 12 \implies \frac{a}{e} = \mathbf{6} \quad \text{(Equation 1)}$$
### D. Eccentricity ($e$) and Semi-transverse Axis ($a$)
We use the relationship between $a, c,$ and the eccentricity $e$:
$$c = a \cdot e$$
Substituting $c=12$:
$$12 = a \cdot e \quad \text{(Equation 2)}$$
Now, we solve the system of equations (1) and (2) for $a$ and $e$:
1. $a = 6e$ (from Eq. 1)
2. $12 = (6e)e$ (substitute into Eq. 2)
$$12 = 6e^2$$
$$e^2 = 2 \implies e = \sqrt{2}$$
Now find $a$:
$$a = 6e = 6\sqrt{2}$$
$$a^2 = (6\sqrt{2})^2 = 36 \times 2 = \mathbf{72}$$
### E. Semi-conjugate Axis ($b^2$)
Use the standard hyperbolic relation $c^2 = a^2 + b^2$.
$$12^2 = 72 + b^2$$
$$144 = 72 + b^2$$
$$b^2 = 144 - 72 = \mathbf{72}$$
---
## 2. Standard Form of the Equation
Since the hyperbola is horizontal, the standard form is:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
Substitute $h=1, k=5, a^2=72,$ and $b^2=72$:
$$\frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} = 1$$
---
## 3. General Form of the Equation
To convert to the general form ($Ax^2 + Bx + Cy^2 + Dy + E = 0$), multiply the entire equation by the common denominator, 72:
$$72 \left( \frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} \right) = 1 \times 72$$
$$(x-1)^2 - (y-5)^2 = 72$$
Expand the squared terms:
$$(x^2 - 2x + 1) - (y^2 - 10y + 25) = 72$$
Remove parentheses and rearrange:
$$x^2 - 2x + 1 - y^2 + 10y - 25 = 72$$
$$x^2 - y^2 - 2x + 10y - 24 = 72$$
Move the constant term to the left side:
$$x^2 - y^2 - 2x + 10y - 24 - 72 = 0$$
$$x^2 - y^2 - 2x + 10y - 96 = 0$$
The equation of the hyperbola in general form is:
$$\mathbf{x^2 - y^2 - 2x + 10y - 96 = 0}$$

Question 740852: What is the center and radius of the circle x^2+y^2-8x-10y+32=0 ?
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
What is the center and radius of the circle x^2+y^2-8x-10y+32=0 ?
~~~~~~~~~~~~~~~~~~~~~

x^2 + y^2 - 8x - 10y + 32 = 0, x^2 + y^2 - 8x - 10y = -32. Group the terms with x (one group) and y (another group) (x^2 - 8x) + (y^2 - 10y) = -32. Complete the squares (x^2 - 8x + 16) + (y^2 - 10y + 25) = -32 + 16 + 25 (x-4)^2 + (y-5)^2 = 9. It is the standard equation of the circle of the radius = 3 with the center at (4,5).

Solved.

The answer in the post by @lynnlo is incorrect.

Question 748181: Write an equation of an ellipse with the given characteristics.
Center (0,0), vertical major axis of length 12, minor axis of length 4
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Write an equation of an ellipse with the given characteristics.
Center (0,0), vertical major axis of length 12, minor axis of length 4
~~~~~~~~~~~~~~~~~~~~~~~~~

The major semi-axis is vertical of the length of 12/2 = 6. The minor semi-axis is horizontal of the length of 4/2 = 2. The standard equation of the ellipse is + = 1.

Solved.

Question 619259: Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        In the post by @Theo, the center is defined correctly, while the radius is incorrect.
        I came to fix it.

x^2+y^2+z^2-4x+2y-6z+10 = 0 becomes:
(x^2 - 4x) + (y^2 + 2y) + (z^2 - 6z) = -10 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = -10 + 4 + 1 + 9 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = 4.
center is (2,-1,3).
radius = sqrt(4) = 2.

Solved.

Question 552371: What is the axis of symmetry of y=7(x+1)(x-3) I have tried another website and it stated something about h=x, and h would be the 1 in the first set of parenthesis, correct? Or is there some other method that is more accurate? Thank you.
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
What is the axis of symmetry of y=7(x+1)(x-3) I have tried another website and it stated
something about h=x, and h would be the 1 in the first set of parenthesis, correct?
Or is there some other method that is more accurate? Thank you.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

32 lines in the post by @Theo is a complete waste of time.
True solution is very simple and very short.

The given polynomial describes a parabola opened up.

This polynomial has two roots at x = -1 and x = 3   (look at the factors in parentheses).

It means that the parabola intersects x-axis in points x = -1 and x=3.

The axis of symmetry of this parabola is vertical line and it is half-way between the roots.

Midpoint between the roots is   x = = = 1.

Therefore, equation of the symmetry line of this parabola is x = 1.

It is how this problem should be solved and how the solution should be presented.

It is a standard mantra to pronounce when solving this problem.

It is what your teacher wants to hear from you.

Question 1168977: Conjugate axis along the y-axis, ome vertex at (0,7), asymptotes are 6x-5y+30=0 and 6x+5y-30=0. Need help finding the Equation of this hyperbola. How?

Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Conjugate axis along the y-axis, one vertex at (0,7), asymptotes are 6x-5y+30=0 and 6x+5y-30=0.
Need help finding the Equation of this hyperbola. How?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

            Regarding this post, I have two notices.

First notice is that the problem's formulation is INCORRECT and SELF-CONTRADICTORY.
It describes a situation which NEVER may happen in reality.

Indeed, the problem says that conjugate axis of the hyperbola is along y-axis,
one vertex is (0,7) and the asymptotes are 6x-5y+30 = 0 and 6x+5y-30 = 0.

From equations, the center of the hyperbola is (x,y) = (0,6).

Together with the information about " one vertex ", it means that transverse axis is vertical y-axis,
and it contradicts that the conjugate axis is y-axis, as stated in the post.

            It disproves the problem, kills it to the death and ruins it into dust.

Second notice is that the solution in the post by @CPhill is TOTALLY and FATALLY incorrect,
starting from his first two lines, where he writes an equation of the hyperbola in WRONG FORM.

So, the problem itself is , as well as its " solution " produced and presented by @CPhill.

For the peace in your mind, my dear reader, ignore both the problem and its " solution " by @CPhill.

            A right place for such compositions is a trash bin.

Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
Solution:
The equation of a hyperbola with a conjugate axis along the y-axis has the standard form:
$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
where $(h, k)$ is the center of the hyperbola, $2a$ is the length of the transverse axis, and $2b$ is the length of the conjugate axis.
The asymptotes of this hyperbola are given by:
$y - k = \pm \frac{a}{b}(x - h)$
We are given the equations of the asymptotes as $6x - 5y + 30 = 0$ and $6x + 5y - 30 = 0$. We can rewrite these in the slope-intercept form ($y = mx + c$):
$5y = 6x + 30 \implies y = \frac{6}{5}x + 6$
$5y = -6x + 30 \implies y = -\frac{6}{5}x + 6$
The slopes of the asymptotes are $\pm \frac{6}{5}$. Therefore, $\frac{a}{b} = \frac{6}{5}$.
The intersection of the asymptotes gives the center of the hyperbola. We can solve the system of equations:
$y = \frac{6}{5}x + 6$
$y = -\frac{6}{5}x + 6$
Setting them equal: $\frac{6}{5}x + 6 = -\frac{6}{5}x + 6 \implies \frac{12}{5}x = 0 \implies x = 0$.
Substituting $x=0$ into either equation gives $y = 6$.
So, the center of the hyperbola is $(h, k) = (0, 6)$.
We are given that one vertex is at $(0, 7)$. For a hyperbola with a vertical transverse axis and center $(0, 6)$, the vertices are at $(0, k \pm a) = (0, 6 \pm a)$.
Since one vertex is at $(0, 7)$, we have $6 + a = 7$ or $6 - a = 7$.
If $6 + a = 7$, then $a = 1$.
If $6 - a = 7$, then $a = -1$, which is not possible for the length $a$.
Thus, $a = 1$, and $a^2 = 1$.
Now we use the relationship $\frac{a}{b} = \frac{6}{5}$ with $a = 1$:
$\frac{1}{b} = \frac{6}{5} \implies b = \frac{5}{6}$.
So, $b^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$.
The equation of the hyperbola is:
$\frac{(y - 6)^2}{a^2} - \frac{(x - 0)^2}{b^2} = 1$
$\frac{(y - 6)^2}{1} - \frac{x^2}{\frac{25}{36}} = 1$
$(y - 6)^2 - \frac{36x^2}{25} = 1$
Final Answer: The final answer is $\boxed{(y - 6)^2 - \frac{36x^2}{25} = 1}$

Question 1209794: (x+4)^2/9+(y+3)^2 /4 =1

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.

It looks like the Artificial Intelligence (represented by @CPhill) reached a dead end and cannot answer correctly.

Meanwhile, the answer is simple:

This equation represents an ellipse with the center at the point (-4,-3)
with a major semi-axis of = 3 units (horizontal) and a minor semi-axis of = 2 units (vertical).

Question 1167285: A router is in the library of the university which is 1.5 years away from a student lounge. The range of the router, which services the WIFI connection of the university, has a diameter of 3.5 meters, with it as the center. Can students staying in the lounge connect to the said WIFI?
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A router is in the library of the university which is 1.5 away from a student lounge.
The range of the router, which services the WIFI connection of the university, has a diameter of 3.5 meters,
with it as the center. Can students staying in the lounge connect to the said WIFI?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Yes, they can, if they are located closer to the router than = 1.75 meters.

Question 1167416: The tower stands 180 meters tall. The diameter of the top is 75 meters. At their closest, the sides of the tower are 60 meters apart. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola is the origin. Round of final answer to the nearest whole number.
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
The tower stands 180 meters tall. The diameter of the top is 75 meters. At their closest,
the sides of the tower are 60 meters apart. Find the equation of the hyperbola that models
the sides of the cooling tower. Assume that the center of the hyperbola is the origin.
Round of final answer to the nearest whole number.
~~~~~~~~~~~~~~~~~~~~~~~~~~

The problem in the post is posed INCORRECTLY.

In the form as it is presented in the post, the problem CAN NOT be solved,
since the given input information is not sufficient.

As the problem is presented, it is clear that the problem's creator is absolutely
(i.e. TOTALLY) illiterate in the subject.

In this form, this task is simply a misunderstanding,
and its appearance on the forum is an unfortunate accident.

In order for the equation of a hyperbola could be restored, the positions (= the coordinates)
of its three points should be given. But in this post, this info is not provided.

It could be solved, if the height of the narrowest part be given; but this info is MISSED in the problem.

The lowest possible score to the problem's creator for his or her unsatisfactory job.

Question 1167713: 1. Kepler observed that Pluto orbits the sun in an elliptic motion. With the
sun at one focus, nearest that Pluto gets to the Sun is 4,400,000,000 km, and the farthest that it goes to the Sun is 7,400,000,000 km.
a) Assuming that the center of Pluto’s orbit is at (0, 0), find the equation
that models Pluto’s orbit.
b) If an unidentified planet is located in the other focus, how far is this
planet from the Sun?
Found 2 solutions by ikleyn, htmentor:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
1. Kepler observed that Pluto orbits the sun in an elliptic motion. With the
sun at one focus, nearest that Pluto gets to the Sun is 4,400,000,000 km, and the farthest that it goes to the Sun is 7,400,000,000 km.
a) Assuming that the center of Pluto’s orbit is at (0, 0), find the equation
that models Pluto’s orbit.
b) If an unidentified planet is located in the other focus, how far is this
planet from the Sun?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

The problem's formulation is MONSTROUS.

        To part (a)

Kepler lived 1571 - 1630.

Pluto was discovered in the year 1930.

So, Kepler (as a person) could not observe Pluto and did not know about Pluto's existence.

        To part (b)

If another planet is " placed " in the other focus (in the standard meaning of the word " placed "),
then, according to Physics, this planet will fall to the Sun.

So, this formulation in (b) is MONSTROUS, too.

Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website!
The equation for an ellipse centered at the origin is x^2/a^2 + y^2/b^2 = 1, where a, b are the semi-major and semi-minor axes, respectively.
The sun is located at one focus with coordinates (-c,0), with c^2 = a^2 - b^2.
For simplicity, we will express the distances in billions of miles.
The distance of the furthest approach is 7.4 = c + a, and the closest
approach is 4.4 = a - c. Solving for a gives 2a = 11.8 -> a = 5.9.
Solving for c gives 2c = 3 -> c = 1.5.
b^2 = a^2 - c^2 -> b^2 = 5.9^2 - 1.5^2 = 32.56.
Thus, the equation is x^2/34.81 + y^2/32.56 = 1.
If another planet is placed at the other focus, the distance between them
will be 2c = 3.

Question 1168412: The Bayonne bridge connects Staten Island, New York to New Jersey. It has an arch in the shape of parabola that opens downward. Write an equation of the parabola to model the arch, assuming that the origin is at the surface of the water. 325. ft and 1675ft
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
Let's break down this problem and find the equation of the parabola.
**1. Understand the Problem**
* The arch of the Bayonne Bridge is in the shape of a downward-opening parabola.
* The origin (0, 0) is at the surface of the water.
* We are given two key dimensions:
* The width of the arch at the water's surface is 1675 feet.
* The height of the arch is 325 feet.
**2. Set Up the Vertex**
* Since the parabola opens downward and the origin is at the water's surface, the vertex of the parabola will be at the highest point of the arch.
* The vertex will be located at (1675/2, 325) = (837.5, 325).
**3. General Equation of a Parabola**
The general equation of a parabola with a vertical axis of symmetry is:
y = a(x - h)² + k
Where:
* (h, k) is the vertex of the parabola.
* 'a' determines the direction and width of the parabola.
**4. Plug in the Vertex**
* We know the vertex is (837.5, 325), so h = 837.5 and k = 325.
* The equation becomes:
y = a(x - 837.5)² + 325
**5. Find 'a'**
* We know the parabola passes through the origin (0, 0). Plug in x = 0 and y = 0:
0 = a(0 - 837.5)² + 325
0 = a(837.5)² + 325
-325 = a(837.5)²
a = -325 / (837.5)²
a = -325 / 701406.25
a ≈ -0.00046335
**6. Write the Equation**
* Substitute the value of 'a' back into the equation:
y = -0.00046335(x - 837.5)² + 325
**7. Simplify (Optional)**
* If needed, we can round 'a' to a more manageable value:
y ≈ -0.000463(x - 837.5)² + 325
**Therefore, the equation of the parabola that models the arch of the Bayonne Bridge is approximately y = -0.000463(x - 837.5)² + 325.**

Question 1165099: A cable hangs in the form of a branch of a hyperbola between two poles that are 20 meters apart. The poles
are 70 meters high and the cable has a sag of 2 meters midway between the poles. Find the height of the cable
at a point 3 meters from one of the poles.
PLEASE HELP ME HUHUHU. THANK YOU SO MUCH!!
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.

Usually, in such problem a shape of a parabola is considered - not a shape of hyperbola.

Therefore, throw your problem to the closest garbage bin, where its natural place is.

Let will not multiply the amount of gibberish in the Internet.

Question 1168931: Find the (standard) equation of the hyperbola whose foci are 𝐹1(-3,0) and 𝐹2(3,0),
such that for any point on it, the absolute value of the difference of its distances
from the foci is 3.
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Find the (standard) equation of the hyperbola whose foci are 𝐹1(-3,0) and 𝐹2(3,0),
such that for any point on it, the absolute value of the difference of its distances from the foci is 3.
~~~~~~~~~~~~~~~~~~~~~~~

            The solution by @MathLover1 is INCORRECT.
            I came to bring a correct solution.

Since the foci are F1(-3,0) and F2(3,0), we conclude that - the major axis is horizontal y= 0 and coincides with x-axis; - the center is at (0,0), half way between the foci; - the distance between the foci is 3 - (-3) = 6 units and the distance from the center to the foci c = 6/2 = 3 units. So, the general equation of the hyperbola is - = 1, where a > 0 is the horizontal transfer semi-axis. Now consider this condition "for any point on it, the absolute value of the difference of its distances from the foci is 3.". We know that 'a' is the distance along x-axis from the center (0,0) to any of the two vertices of the hyperbola. Then the points (-a,0) and (a,0) are the points on this hyperbola. For point (a,0), the distance from F1 is (3-a); the distance from F2 is (a+3). The difference of these distance must be 3, according to the condition. So, we write this equation (a+3) - (3-a) = 3, 2a = 3, a = 3/2 = 1.5. So, the vetices are at (-1.5,0) and (1.5,0), and 'a', as we found, is 1.5. Then from equation a^2 + b^2 = c^2 we can find 'b^2' 1.5^2 + b^2 = 3^2 b^2 = 3^2 - 1.5^2 = 9 - 2.25 = 6.75. Hence, the standard equation of this hyperbola is - = 1. <<<---=== ANSWER

Solved correctly.

Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

given:
foci are F1(,) and F2(,),
such that for any point on it, the absolute value of the difference of its distances from the foci is

The standard form of a hyperbola with a horizontal transverse axis is

where (,) is the center of the hyperbola
is the distance from the center to a vertex
is the distance from the center to a co-vertex

The center of the hyperbola is the midpoint of the line segment between the foci, which is

so, the center of the hyperbola is at (,)
The distance between the foci is , so =>
The distance from each focus to the center is
.
using find

So, the equation of the hyperbola is:

since , this is a valid hyperbola
The foci and vertices coincide, so the hyperbola degenerates into a line.
The hyperbola with the given parameters doesn't exist.

Question 1168800: Write an equation in standard form of the parabola that has the same shape as the graph of
f(x)= 4x^2 or
g(x)=−4x2, but with the given maximum or minimum.
Maximum=7 at x=-2

Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

Write an equation in standard form of the parabola that has the same shape as the graph of
or , but with the given maximum or minimum.

given:
at => vertex is at (,)=(,) =>,
To move a parabola units to the right, you would substitute for , and to move up ad units in the corresponding quadratic equation
so, using
where ,
The equation in a standard form of the parabola that has a maximum at at is

Question 1168877: Find the length of latus rectum (LR) is 10, vertex at the origin; parabola opens to the right.
Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!
given: the length of latus rectum (LR) is
The length of latus rectum for a standard equation of a parabola is equal to
=>
then and foci is at (,)
and your equation is:

Question 1168920: Find the equation of the ellipse if the endpoints of minor axis are (1,3) and (1,-1) with a focus at (-1,1).Express your answer in general form.

Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

The equation of an ellipse can be given as,

where
'' represents the semi-major axis (half of the length of the major axis)
'' represents the semi-minor axis (half of the length of the minor axis)
‘’ represents the coordinate of the center
‘’ represents the coordinate of the center

if the endpoints of minor axis are (,) and (,) , minor axis length is equal to the distance between them which is
half of the length of the minor axis is
=> the center mast be half way between endpoints, and it is at (,)

a focus at (,)

the distance between foci (,) and center (,)
focus is units from the center, so

using the Pythagorean fact of all ellipses

so, your formula is

Question 1168978: Foci at (-1,4) and (7,4), transverse axis has the length 8/3. Whats the equation
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
Solution:
The given information indicates that the conic section is a hyperbola because we are given foci and a transverse axis length.
The foci are at $F_1(-1, 4)$ and $F_2(7, 4)$.
The transverse axis has a length of $\frac{8}{3}$.
From the foci, we can determine the center of the hyperbola as the midpoint of the foci:
Center $(h, k) = \left( \frac{-1 + 7}{2}, \frac{4 + 4}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)$.
The distance between the foci is $2c$.
$2c = \sqrt{(7 - (-1))^2 + (4 - 4)^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8$.
So, $c = 4$.
The length of the transverse axis is $2a$. We are given $2a = \frac{8}{3}$, so $a = \frac{4}{3}$.
Since the y-coordinates of the foci are the same, the transverse axis is horizontal. The standard form of the equation of a hyperbola with a horizontal transverse axis and center $(h, k)$ is:
$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$
We need to find $b^2$. We use the relationship $c^2 = a^2 + b^2$ for a hyperbola.
$4^2 = \left(\frac{4}{3}\right)^2 + b^2$
$16 = \frac{16}{9} + b^2$
$b^2 = 16 - \frac{16}{9} = \frac{144 - 16}{9} = \frac{128}{9}$.
Now, substitute the values of $h=3$, $k=4$, $a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$, and $b^2 = \frac{128}{9}$ into the standard form:
$\frac{(x - 3)^2}{\frac{16}{9}} - \frac{(y - 4)^2}{\frac{128}{9}} = 1$
Multiply the numerators by $\frac{9}{16}$ and $\frac{9}{128}$ respectively:
$\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1$.
Final Answer: The final answer is $\boxed{\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1}$

Question 1169000: find the equation of the ellipse if the vertex is at (5,1) and foci at (-2,1) and (4,1). express your answer in general form.
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website!
Solution:
The given information about the ellipse is:
Vertex: $(5, 1)$
Foci: $(-2, 1)$ and $(4, 1)$
Since the y-coordinates of the vertex and the foci are the same, the major axis of the ellipse is horizontal.
The center of the ellipse is the midpoint of the foci:
Center $(h, k) = \left( \frac{-2 + 4}{2}, \frac{1 + 1}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$
The distance between the center and a focus is $c$. Using the focus at $(4, 1)$:
$c = \sqrt{(4 - 1)^2 + (1 - 1)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$
The distance between the center and a vertex is $a$. Using the vertex at $(5, 1)$:
$a = \sqrt{(5 - 1)^2 + (1 - 1)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$
We know the relationship between $a$, $b$, and $c$ for an ellipse: $c^2 = a^2 - b^2$.
Substituting the values of $a$ and $c$:
$3^2 = 4^2 - b^2$
$9 = 16 - b^2$
$b^2 = 16 - 9 = 7$
The standard form of the equation of a horizontal ellipse with center $(h, k)$ is:
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
Substituting the values of $h=1$, $k=1$, $a^2 = 16$, and $b^2 = 7$:
$\frac{(x - 1)^2}{16} + \frac{(y - 1)^2}{7} = 1$
To express the equation in general form, we eliminate the denominators and expand:
$7(x - 1)^2 + 16(y - 1)^2 = 16 \times 7$
$7(x^2 - 2x + 1) + 16(y^2 - 2y + 1) = 112$
$7x^2 - 14x + 7 + 16y^2 - 32y + 16 = 112$
$7x^2 + 16y^2 - 14x - 32y + 23 = 112$
$7x^2 + 16y^2 - 14x - 32y + 23 - 112 = 0$
$7x^2 + 16y^2 - 14x - 32y - 89 = 0$
Final Answer: The final answer is $\boxed{7x^2 + 16y^2 - 14x - 32y - 89 = 0}$

Question 1169047: At this point, you are now ready to take the summative assessment for learning plan 3. Place your answers on a whole sheet of paper. Show your complete solution. (20 points)
1. Write the standard form of the equation of the hyperbola with the following characteristics:
Center at (2,3)
Vertices at (-1,3) and (5,3)
Covertices at (2,-2) and (2,8)
Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!
recal:
Coordinates of the center: (, ).
Coordinates of vertices: (, ) and (,)
Co-vertices correspond to , the ” minor semi-axis length”, and coordinates of co-vertices:
(,) and (,).
Foci have coordinates (,) and (,). The value of is given as, .
equation:

given:
Center at (,)=(,)
Thus, , .
Vertices at (,) and (,)
The distance between the vertices is . We can find this distance by subtracting the x-coordinates of the vertices:

so far your equation is:

Covertices at (,)and (,)
since covertices at (,) and (,), we have
(,) =(,) => => =>

so, your equation is:

Question 1169036: find the equation of the ellipse if the endpoints of minor axis are (1,3) and (1,-1) with a focus at (-1,1).
Answer by mccravyedwin(421)   (Show Source):
You can put this solution on YOUR website!

Notice that on the first page this problem reads this way: Quadratic-relations-and-conic-sections/1169036 (2020-11-05 01:23:06): See the " 2020-11-05 01:23:06 "? This means that this problem was posted by a student 4 1/2 years ago on November 5, 2020 at 1:23:06 AM. The student who posted this is no doubt long gone from the class he or she was taking back in 2020, and is no longer a student. Back in 2020 there were still lots of algebra and trig problems being posted. There weren't enough tutors on here, and many problems like this one scrolled off before the tutors got around to solving them because there were so many. But then the schools learned that kids can take basic statistics with almost no knowledge of second year algebra, no geometry, or trig. First year algebra is enough mathematics for basic statistics, especially now since it's mostly all done on computers. So apparently, now most high school students take statistics instead of trig and college algebra. That's why practically all posts on here nowadays are statistics problems. OK, I'll shut up and do this problem posted back in 2020. The former students getting solutions in their emails to problems they posted back in 2020 are probably having a good laugh! J ---------------------- Find the equation of the ellipse if the endpoints of minor axis are (1,3) and (1,-1) with a focus at (-1,1). First draw what is given, the minor axis, and plot a focus at (-1,1). Since the minor axis is vertical, we know that the major axis is horizontal, and the ellipse is like the cross section of an egg sitting on a table, [and not like the number 0, when the minor axis is horizontal and the major axis is vertical.] So the length of the semi-major axis ' a ' goes under the term with x, and the length of the semi-minor axis ' b ' goes under the term with y. So the form of the equation is Where the center is the point (h,k). Next, we draw the center, which is the midpoint of the minor axis. We can tell the center is the point (1,1). so we know that the center (h,k) = (1,1). So we have everything but ' a '. So we can write this much of the equation: No we calculate ' a '. The value " c " is the distance from the center to a focus. We can tell that the distance from the center (0,0) to a focus (-1,1) is 2 units, so the value of c = 2. Then we use the Pythagorean fact of all ellipses, which is where a is the length of the semi-major axis, b is the length of the semi-minor axis, and c is the distance from the center to a focus. Since the minor axis that we drew is 4 units long, the semi-minor axis, b, is half that, or b=2. So we substitute b=2 and c=2 into So now we know the equation is or <----ANSWER! But let's finish drawing the ellipse, because you will be asked to do that: which is about 2.8. So draw the major axis, which goes from about 2.8 units left of the center, to about 2.8 units right of the center. Now we can sketch in the ellipse. And we might as well plot the other focus which is 2 units on the right of the center. Edwin

Question 1167082: a flashlight is shaped like a paraboloid so that if its bulb is places at then focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the x-axis. if the paraboloid has a depth of 1.8 inches and the diameter on its surface is 6 inches, how far should the light source be placed from the vertex
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
a flashlight is shaped like a paraboloid so that if its bulb is places at then focus,
the light rays from the bulb will then bounce off the surface in a focused direction
that is parallel to the x-axis. if the paraboloid has a depth of 1.8 inches
and the diameter on its surface is 6 inches, how far should the light source be placed from the vertex
~~~~~~~~~~~~~~~~~~~~~

        It is a standard problem of this kind. The feature is
        that in this problem the symmetry line is parallel to x-axis,
        while usually in such problems the symmetry line is parallel to y-axis.
        So, I will adapt a standard solution to this case.

For solving such problems, write an equation of the parabola in the cross-section in the form x = . (1) The advantage of writing in this form is the fact that then "p" is the distance from the parabola vertex to its focus. So, now our task is to find value of "p" from the given data. The fact that "the paraboloid has a depth of 1.8 inches and the diameter on its surface is 6 inches" means that x = 1.8 meters at y = 6/2 = 3 inches. So, we substitute this data into equation (1), and we get 1.8 = . It gives p = = 1.25 = 1 inches. Thus, the light source should be placed 1 inches from the vertex. ANSWER

Solved.

Question 1167899: Please answer this question, I tried but I don't understand it.
Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex. For reference, the first considers a parabolic string with a base 10 ft. and a focus at 5 ft.from the vertex. How deep is the parabolic solar cooker?
Thank you for your help!
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Please answer this question, I tried but I don't understand it.
Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex.
For reference, the first considers a parabolic string with a base 10 ft. and a focus at 5 ft. from the vertex.
How deep is the parabolic solar cooker?
Thank you for your help!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In this problem, you are given the size 10 ft of the opening of the parabolic solar cooker,
and the distance 5 ft from the vertex to the focus of the paraboloid (same as the place for cooking).

They want you find the depth of the parabolic mirror.

                                 S O L U T I O N

To solve such problems, use an equation of the parabolic cross-section (which is a parabola) in the form y = . (1) In this form, 'p' is the focal distance, i.e. the distance of 5 ft from the vertex to the cooking place; x is half of the opening, i.e. 10/2 = 5 ft; y is the depth of the paraboloid. So, we substitute p = 5 ft and x = 5 ft into equation (1), and we get y = = = ft = 1 ft. At this point, the problem is completely solved. ANSWER. The depth of the paraboloid is 1 ft.

Solved.

Question 1165710: A cable hangs in the form of a branch of a hyperbola between two poles that are 20 meters apart. The poles are 70 meters high and the cable has a sag of 2 meters midway between the poles. Find the height of the cable at a point 3 meters from one of the poles.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A cable hangs in the form of a branch of a hyperbola between two poles that are 20 meters apart.
The poles are 70 meters high and the cable has a sag of 2 meters midway between the poles.
Find the height of the cable at a point 3 meters from one of the poles.
~~~~~~~~~~~~~~~~~~~~~~~~~~

Usually/traditionally, in such problems, a parabolic form is considered,
because parabolic form better corresponds to Physics.

Let's not multiply the amount of gibberish in this world.

Question 1165555: A cable hangs in the form of a branch of a hyperbola between two poles that are 20 meters apart. The poles
are 70 meters high and the cable has a sag of 2 meters midway between the poles. Find the height of the cable
at a point 3 meters from one of the poles.
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A cable hangs in the form of a branch of a hyperbola between two poles that are 20 meters apart. The poles
are 70 meters high and the cable has a sag of 2 meters midway between the poles. Find the height of the cable
at a point 3 meters from one of the poles.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Usually/traditionally, in such problems, a parabolic form is considered,
because parabolic form better corresponds to Physics.

Let's not multiply the amount of gibberish in this world.