# Questions on Algebra: Conic sections - ellipse, parabola, hyperbola answered by real tutors!

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Question 994389: Use the following equation: y = –3(x + 8)^2 + 2
The coordinates of the vertex written as (h, k) would be?

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They are (-8,2). You reverse the sign of the constant in the parentheses and take the last constant without changing sign.
This is also
f(x)=-3x^2-48x-190.
The vertex is -b/2a which is 48/-6 =-8
f(x)=-192+384-190=2

Question 993785: for the parabola : y = -x^2 + 36.
Graph of a parabola opening down at the vertex (0,36) crossing the x–axis at (-6,0) and (6,0).

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John

My calculator said it, I believe it, that settles it

Question 993571: find the equation of a circle that passes thru'(6,0) and (24,0) and is tangent to the y-axis

Found 2 solutions by solver91311, anand429:
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The segment of the x-axis bounded by the points (6,0) and (24,0) is a chord of the desired circle. Since the perpendicular bisector of a chord must pass through the center of the circle, we know that the center of the desired circle lies on the line x = 15. We also know, since said line is a bisector of the chord, that the distance from the point (6,0) to the point of intersection of the line x = 15 and the x-axis is 9. Since the circle is tangent to the y-axis, the distance from the y-axis to the center of the circle must be 15, hence the radius of the circle is 15.

Consider the right triangle formed by the radius of the circle that has the point (6,0) as an end point, the portion of the line x = 15 from the center of the circle to the center of the chord segment of the x-axis, and the half-chord from point (6,0) to (15,0). Since this is a right triangle with hypotenuse/short leg ratio of 5:3, the other leg must be in ratio 5:4:3 with the other two sides. Hence the distance from the x-axis to the center of the circle is 12.

Our circle is centered at (15,12) and has a radius of 15.

A circle centered at (h, k) with radius r has an equation:

John

My calculator said it, I believe it, that settles it

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(6,0) and (24,0) both lie on x-axis, hence mid point of these points will lie directly below( or above) centre (Since line joining centre to mid point of a chord is perpendicular to the chord)
Also, since the circle touches the y axis, so distance from origin to this mid point found above will be equal to radius.
Now mid point is ((6+24)/2,(0+0)/2) i.e. (15,0)
Now let distance of centre from mid point found above be p
So, (DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)
i.e.
=> p = 12 or p = -12
So coordinates of centre will become (15,12) or (15,-12)
So, the equation of circle,
or
=> or
(DRAW DIAGRAM TO UNDERSTAND ALL THE STEPS)

Question 993227: find the standard form of the equation of the parabola satisfying the given conditions
focus (2,1) directrix x=-4

Question 992413: Equation of parabola and latus rectum with focus (-8,0) and directrix x=8
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You can derive the equation using this idea:
Equation of Parabola for vertex other than origin

Question 992302: Hyperbola
Determine the center,vertices, foci,conjugat, and the latus rectum of the
(y+3)^2/16 - (x-2)^2/25=1

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....if you compare to you see that ,, ,
since the part of the equation is added, then the center, foci, and vertices will be above and below the center (on a line paralleling the y-axis)
so,
the center is at (,)= (,)
semi-major axis length
semi-minor axis length

since and , the equation tells me that , so , and
the eccentricity is

the vertices and foci are above and below the center,
so the foci are at
(,)=(,) and
(,)= (,)
or, approximately at (, ) and (,)

the vertices are at (,) and (, )
(,) =>(,)
and
(, )=>(,)

The length of the Latus Rectum:
In a hyperbola, it is twice the square of the length of the transverse axis divided by the length of the conjugate axis.
the length of the transverse axis is =>=>
the length of the conjugate axis is =>=>
the Latus Rectum is

Question 992315: directrix has y-intercept 3
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.
This request is fatally incomplete.

Question 992151: x^2/4+y^2/9+z^2/16=1

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.
Thank you for letting us know.

Question 992106: Find the equation of the parabola with focus (5,4) and directrix x = 3

Question 991825: Please help me with this! Find the centre, vertices, lenght of latus rectum, foci , eccentricities of the ellipse x^2 + 9y^2 + 4x -18y - 23 = 0
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.
x^2 + 9y^2 + 4x -18y - 23 = 0
----------------------------------------------

Complete the squares:

= = .

So,  you have the equation

+ = ,

or,  in the canonical form,

+ = .

It is a canonical equation of an ellipse.

Its center is at the point  (x,y) = (-2,1).  Semi-major and semi-minor axes are  6  and  2,  respectively:  a=6  and  b=2.
Semi-major axis is the distance from the ellipse' center to its vertices.

Linear eccentricity is   = = .  It is the distance from the ellipse's center to its foci.

Next,  take the textbook and determine the rest ellipse's characteristics you need.

Question 991566: How do you the parabola y=4x^2-12x+5 into vertex form and intercept form?
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Complete the square to put into vertex form,

Then factor to get to intercept form,

.
.
.
.

Question 991544: What is the vertex form of the equation y=4(x+1)(x-1)?
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Since the zeros are at and , then the vertex will be at .
When ,.
(0,-4)

Or multiplying it out.

(0,-4)

Question 991186: 4x^2+9y^2=25 at (-2,-1)
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Do you have a question?

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Quadratic equation in two variables, independant variable leading term has negative coefficient. This is a parabola opening upward, having vertex at a minimum.

This lesson will help with to find the vertex and the roots (or also called Zeros) of the parabola:
If unclear AFTER studying that a couple of times carefully, say so.

Question 988025: Question Part
Points
Submissions Used
Use the quadratic formula to find any x-intercepts of the parabola. (If an answer does not exist, enter DNE.)
y = 2x^2 − 10x + 5

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At the x-intercepts, , so

----------------------------

and

------------------------

close enough- you can check the other x-intercept

Question 987973: (Ellipse)
25x^2 + 64y^2 - 400x - 600y=0
Find its standard equation, foci, vertices and extemeties of the minor axis.

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.........complete square

.........since ...

....multiply by
....divide by

now we have a standard form and we know that:
=>
=>
we can calculate using formula =>

we also know that

=>so, the center is at (, ), approximately at (, )
foci:(, ) or (, ), approximately at (, ) or (, )
vertices: (, ) or (, ), approximately at (, ) or (, )
semi-major axis length: ~~
semi-minor axis length: (~~
eccentricity | ~~

Question 987549: Test for symmetry
4x^2=+y^2=64

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Too many equal signs.

Question 987347: what are the coordinates of the center of the ellipse graphed by the equation below? (x+14)^2/6^2+(y+17)^2/5^2=1?
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.
For an ellipse:

In this form, (h,k) is the center.
.

h=-14 and k=-17, so center is at (-14,-17)

Question 986810: Put the equation in standard form for a hyperbola.
4x2−9y2−24x−54y=81

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term to complete the square for each of the quadratic factors INSIDE the grouping symbols is .
-----you need to clearly understand this step;

---
Explanations for Completing The Square:
-
Lesson on how to solve, but includes how to choose the term to Complete The Square, and how to perform the process
-
Visual explanation of what Completing The Square means

Question 986493: Sketch the graph of 9x2 – 16y2 = 144. Label any intercepts and asymptotes.

Question 986364: How do I algebraically determine where these two conic sections intersect?
4x^2+16y^2=64
2x-y^2=-4

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"It's deja vu all over again." Yogi Berra

Question 986363: How can I algebraically find out where these two conic sections intersect?
4x^2+16y^2=64
2x-y^2=-4

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How can I algebraically find out where these two conic sections intersect?
4x^2+16y^2=64
2x-y^2=-4
=====================
I would solve the 2nd eqn for x, then sub for x in the 1st eqn.

Question 985894: How do you find the focus and directrix of the parabola y=x^2-4x+4 and graph it?
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first write your equation in vertex form : where and are and coordinates of the vertex
......complete square; here you have
=> , and
vertex is at:(,)
=> parabola opens upwards
then the focal distance , in this case and
focus | (, )
semi-axis length |
directrix |

Question 985669: a.

What is the maximum value of this expression?
b.

For which values of y will x be positive?

Found 2 solutions by Alan3354, macston:
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a.

What is the maximum value of this expression?
Without calculus:
f(x) = -5x^2 + 12x
The max is the vertex at x = -b/2a = -12/-10 = 1.2
f(1.2) = 14.4 - 7.2
= 7.2
======================
b.

For which values of y will x be positive?
Find where y = 0
 Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=33 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 0.593070330817254, -0.843070330817254. Here's your graph:

================
x = 0 at those 2 points, x1 & x2
y is positive for x < x1 and x > x2

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a.

Maximum occurs when first derivative=0
Set=0

= Maximum value when x=6/5.
.

=
.
Please limit to one question per post.

Question 985675:
For which values of x will y be negative?

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For which values of x will y be negative?
================
Find the 2 zeroes:
y = (x-1)*(x-5)
y = 0 at x = 1 and at x = 5
y < 0 for 1 < x < 5

Question 985667:
What is the smallest positive integer for b that will force this parabola to have two x-intercepts?

Answer by Edwin McCravy(13211)   (Show Source):
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That will be when the discriminant is positive:

b²-4ac = b²-4(6)(11) > 0
b²-264 > 0
b² > 264

Taking positive square roots:

b > 16.24807681

Smallest positive integer is the next higher positive integer, b=17.

When b=16, no x-intercepts:

When b=16.24807681, 1 x-intercept:

When b=17, 2 x-intercepts:

Edwin

Question 985482: For the circle and the line , determine the exact values of the gradient so that the line:
a) is a tangent to the circle
b) intersects the circle in two places
c) does not intersect the circle

Found 3 solutions by ikleyn, solver91311, KMST:
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To find the intersection points of the circle and the straight line,  substitute   =   into the equation   = .  You will get
+ = ,

+ = ,

= ,

= .

Next,  apply the quadratic formula to find the roots

= ,

where    is the discriminant   = = = .

Now everything is determined by the discriminant.

If   > ,  then there are two roots.  Correspondingly,  there are two intersection points.
In opposite,  if there are two intersection points,  then there are two roots, hence, > .

If   = ,  then there is only one root.  It means that the straight line tangents the circle.
And in opposite,  if the straight line tangents the circle, then there is only one root,  hence,   = .

Finally,   < <-----> there are no intersection points.

Thus you need to solve this critical equation

= .

It gives

= +/- .

So,  if   = +/- ,  then the straight line tangents the circle.

If   > ,  then  >  and there are two intersection points.

If   < < ,  then <  and there are no intersection points.

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KMSTs solution is more general, and therefore better, IMHO, than what I am about to show you. However, this is a slightly different way to look at this particular problem and may give you some additional insight.

Note that a tangent to a circle is perforce perpendicular to a radius. An examination of the diagram shows us that a tangent as described, the perpendicular radius, and that segment of the -axis between the origin and the -intercept of the desired line form a right triangle. Since the measure of the radius of the circle which forms the short leg of the right triangle is 2 and the measure of the -axis segment that forms the hypotenuse of the right triangle measures 4, we have a 30-60-90 right triangle, where the angle between the radius and the axis is the 60 degree angle. See figure below:

.

That means that the angle between the radius ray and the -axis is 30 degrees. A quick examination of a unit circle and using a multiplier of 2 to match the radius of the given circle, we see readily that the point of tangency in QI is .

Using the slope formula, the slope of the line that passes through the points and is .

Symmetry gives us a second point of tangency in QII, which is simply and then the slope of the line through the second point is just the additive inverse of the slope of the first equation.

From there it should be clear that any slope in the open interval will produce a secant line.

Further, any slope in either open interval or will produce a line that does not intersect the circle.

John

My calculator said it, I believe it, that settles it

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gives us the intersection points of circle and line.
--->--->--->
A quadratic equation of the form
has no real solutions if .
It has one real solution if , and
it has two real solutions if .
For the quadratic equation , ,so
.
The zeros of are the zeros of , given by
and .
For those values, the line is tangent to the circle.

In between those values of ,
for ,
is negative, and the line does not intersect the circle.
Otherwise, for ,
, and the line intersects the circle in two places.

Identify the vertex, focus,and directrix of equation:
y=1/12x^2

Found 2 solutions by macston, josgarithmetic:
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.

where (h,k) is the vertex and p is distance from vertex to focus
.
Given equation:

So vertex=(h,k)=(0,0)
Focus is at p from vertex, +1/4 from vertex=(0,1/4)
Directrix is -p from vertex, y=-(1/4)
.

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Deriving Equation for Parabola

Your parabola has its vertex as a minimum, and parabola opens upward.

Question 984794: Find the fifth roots of 32(cos 280° + i sin 280°).
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i) Let z = 32(cos 280° + i sin 280°) = 32{cos(k*360° + 280°) + isin(k*360° + 280°)},
where k is an integer.
ii) ==> z^(1/5) = [32{cos(k*360° + 280°) + isin(k*360° + 280°)}]^(1/5)
==> z^(1/5) = 2[cos{(k*360° + 280°)/5} + isin{(k*360° + 280°)/5}]
[Application of De-Moievere's theorem]
==> z^(1/5) = 2[cos(72k + 56) + isin(72k + 56)]
We can now get the 5 roots, by assigning k = 0, 1, 2, 3 & 4
When k = 0, 1st root = 2(cos 56° + isin 56°)
When k = 1, 2nd root = 2(cos 128° + isin 128°)
When k = 2, 3rd root = 2(cos 200° + isin200°)
When k = 3, 4th root = 2(cos 272° + isin 272°)
When k = 4, 5th root = 2(cos 344° + isin344°)

Question 984603: Find an equation of the parabola described then graph the parabola. Focus(-2,1); Vertex(-3,1)
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Knowing vertex and focus, you can find the directrix.; the distance between vertex and the focus is the same distance between the vertex and the nearest point on the directrix.

The number p is the distance between the vertex and the focus. Knowing p and the vertex, you can write an equation for the parabola directly.

Your parabola opens toward the left, so you will have a form like .

Either or both of these video presentations will help:
Equation of parabola, vertex at origin - Derivation
-
Equation of parabola, vertex not at origin, focus and directrix given - general

Question 984535: I have two limit questions:
Find the derivative of f(x) = 2/x at x = -2.
Find the derivative of f(x) = -9/x at x = -8.

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I'll do the first problem to get you started.

-------------------------------------------------------

-------------------------------------------------------------------------------

Therefore, the derivative of f(x) is

Now plug in x = -2 to get

I'll let you do the other problem.

Question 984414: solve the system.
x+y-2z=8
x+3y+2z=14
x-3y-2z=4

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Use Cramer's rule,

.
.
.

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.
.

.
.
.

.
.
.

Question 984409: I have two limit questions:
Find the derivative of f(x) = 5/x at x = -1.
Use graphs and tables to find the limit and identify any vertical asymptotes of lim(x-->7) 1/(x-7)^2

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I already did this one for you, Question 984344. But here it is again.

Re-write:

Then use the power rule

You can evaluate yourself.

As for the other one, here is the graph. You should be able to figure out the rest.

.

John

My calculator said it, I believe it, that settles it

Question 983852: What is the standard equation of the parabola with given: vertices (2,5) parallel to x-axis and passes through (7,1)
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Hi dear,
is the equation of a parabola.
since the parabola pass through the (7,1),
then substitute the point in the equation above.

a=4/5
Also the standard equation of a parabola is given by

where (h,k) is the coordinate of it's vertice.
If a=4/5
(h,k)=(2,5)
substitute the above in the standard equation and get

Question 983937: The height of the water, H, in feet, at a boat dock t hours after 6 a.m. is given by H=8+sin pi/3 t. Find the height of the water at 9a.m.
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9 AM is three hours after 6 AM. Plug in 3 for t and do the arithmetic.

John

My calculator said it, I believe it, that settles it

Question 983860: What is the standard equation of the parabola with given vertices (2,5), parallel axis to x axis and if its passes through (7,1)
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This way of showing generalized standard form if parallel to the x-axis:

You will use the point (7,1) to find a, and also use the given vertex h=2, k=5.

Put in the values into the form:
or if you want, .

Question 983479: Write the equation of an ellipse with vertices at (7, 0) and (-7, 0) and co-vertices at (0, 1) and (0, -1).

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<-->
If vertices and co-vertices are symmetrical with respect to the origin,
then the ellipse is centered at the origin.
The equation of an ellipse centered at the origin is
.
When , ---> are the coordinates of vertices or co-vertices.
When , ---> are the coordinates of vertices or co-vertices.

Question 983482: Write the equation of an ellipse with a major axis of length 8 and co-vertices (0, 3) and (0, -3).

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Major axis length of 8 means a=4. Co-vertices indicate b=3. The a goes with the x.

.

Question 983483: Write the equation of an ellipse with vertices (0, 12) and (0, -12) and co-vertices (2, 0) and (-2, 0).