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

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Question 572346: Please help me solve and graph this equation:
I tried putting it in standard form and it ended up being a horizontal hyperbola. The standard form was
From there I got the center of the hyperbola which is (h,k) or (-2,-4)and I graphed it. Then I looked for the values for a, b, and c. A was or 2.6 as a decimal and B was or 7.9 as a decimal. I graphed A and B as the conjugate and transverse axes. This is where I got confused. The equation for the hyperbola had x on the left and y on the right, meaning it should graph horizontally. But I graphed the axes and the hyperbola came out vertical. Please help me see where I went wrong!

Found 2 solutions by KMST, lwsshak3:
Answer by KMST(600)   (Show Source):
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Here's how I go about it:
--> --> --> --> --> --> --> -->
and dividing both sides by 36

Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
Solve and graph
9x^2-4y^2+18x+32y-91=0
complete the square
9(x^2+2x+1)-4(y^2-8y+16)=91+9-64=36
9(x+1)^2-4(y-4)^2=36
(x+1)^2/4-(y-4)^2/9=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
..
For given equation:
Center:(-1,4)
a^2=4
a=2
..
b^2=9
b=3
..
slope of asymptotes=±b/a=±3/2
Equation of asymptotes:
y=mx+b
y=-3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=-(3*-1)/2+b
b=5/2
Equation:y=-3x/2+5/2
..
y=3x/2+b
solving for b using (x,y) coordinates of center thru which the asymptote goes.
4=(3*-1)/2+b
b=11/2
Equation:y=3x/2+11/2
..
I don't have the means to graph it for you, but you now have the information you need to do it.

Question 572263: can you visualy show me step by step how to find the vertex of the equation y=5(x+5)(x-2)

Answer by nyc_function(2645)   (Show Source):
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y = 5(x+5)(x-2)
y = 5(x^2 + 3x - 10)
y = 5x^2 + 15x - 50
The vertex we need is in the form (x,y).

To find the x-coordinate, use x = -b/2a.

a = 5 and b = 15

x = -15/2(5)

x = -15/10

x = -3/2

To find y, replace x in the above equation with -3/2 and simplify.

y = 5(-3/2)^2 + 15(-3/2) - 50

y = -245/4

The vertex is the pont (-3/2, -245/4).

Question 572049: Write the standard form of the equation of the circle with the given center (-2,-4) and radius of 6
Answer by Theo(2978)   (Show Source):
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standard form of equation of a circle is:
(x-h)^2 + (x-k)^2 = r^2
(h,k) is the center of the circle.
r is the radius.
in your equation, this becomes:
(x+2)^2 + (y+4)^2 = 36
to graph your equation, solve for y to get:
y = +/- sqrt(36 - (x+2)^2) - 4
that graph is shown below:

Question 571627: how do i classify conic sectins n write its equation in standard form?
example:
-2y^2+x-20y-49=0
Answer by KMST(600)   (Show Source):
You can put this solution on YOUR website!
The conic sections you are likely to encounter will have axes of symmetry parallel to the x and y axes, and that will make it easier, because you will not see a term with xy.
Getting to the standard form involves completing squares. Whenever you see a variable and a variable squared, as in
you have to imagine the part of the expression with that variable as part of a perfect square.
In this case
should come to mind.
With that in mind, you start transforming the equation
--> --> --> --> --> --> --> -->
The last equation tells you that it is a parabola with horizontal axis of symmetry , vertex at (-1,-5), and focal distance
It opens to the right. ,--> .
The directrix is the vertical line.
The focus has , so it's the point(-0.875,-5).

Question 571501: a ball is thrown so that it starts out horizontally. it drops 0.3 m in going and 3 m horizontally. (a) how far does it drop in going 6 m horizontally? (2) how far horizontally does it go in dropping 2.5 m?
Answer by KMST(600)   (Show Source):
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Physics textbooks show that the trajectory of a ball that rolls of a table is a parabola with drop proportional to the square of the horizontal distance.
Let x be the horizontal distance covered as the ball is dropping (in m).
Let y be the drop distance (in m).

Substituting y=0.3 and x=3
--> --> -->
(a) for x=6, --> -->
The ball drops 1.2 m as it moves 6 m horizontally.
(b) for y=2.5 --> --> --> = approximately 8.7
The ball has moved 8.7 m horizontally as it dropped 2.5 m.

Question 570551: Solve the system of equations
y^2-3x^2=6
y=2x-1
Answer by lwsshak3(2927)   (Show Source):
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Solve the system of equations
y^2-3x^2=6
y=2x-1
**
substitute (2x-1) for y in first equation
y^2-3x^2=6
(2x-1)^2-3x^2=6
4x^2-4x+1-3x^2=6
x^2-4x-5=0
(x-5)(x+1)=0
x=5
or
x=-1

Question 570149: A diameter is the segment from (-6,-4) to (3,5) ... Find the equation of the circle.
Answer by lwsshak3(2927)   (Show Source):
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A diameter is the segment from (-6,-4) to (3,5) ... Find the equation of the circle.
**
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
Finding mid points of x and y coordinates of given line segment end points:
x=(3-6)/2=-3/2=-1.5
y=(5-4)/2=12=.5
center of circle: (-1.5,0.5)
..
Finding diameter of circle using distance formula:
d^2=(3+6)^2+(5+4)^2=9^2+9^2=81+81=162
d=√162
radius=d/2=√162/2≈6.36
radius^2=40.5
Equation of circle:
(x+1.5)^2+(y-0.5)^2=40.5

Question 571013: Write the equation of an ellipse with foci (0, 18) and (0, —12) and major axis length 34. Please Hurry!
Answer by lwsshak3(2927)   (Show Source):
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Write the equation of an ellipse with foci (0, 18) and (0, —12) and major axis length 34.
**
Standard form of equation for an ellipse with vertical major axis:
(x-h)^2/b^2+(y-k)2/a^2=1,a>b, (h,k) being the (x,y) coordinates of the center.
for given equation:
center: (0,3)
length of major axis=34=2a
a=17
a^2=289
..
c=15
c^2=225
..
c^2=a^2-b^2
b2=a2-c2=289-225=64
..
Equation of given ellipse:
(x)^2/64+(y-3)^2/289=1

Question 570987: If the quadratic equation f(x)=x^2+2x-2 were translated up 5 units and to the right 4 units, then what would be the zeros of the resulting quadratic function? Explain your solution.
Answer by lwsshak3(2927)   (Show Source):
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If the quadratic equation f(x)=x^2+2x-2 were translated up 5 units and to the right 4 units, then what would be the zeros of the resulting quadratic function? Explain your solution.
***
Translating a function up or down is graphically in effect moving the entire curve up or down, that is, f(x) goes up or down. In this case, moving f(x) 5 unit up looks like this: f(x)+5=(x^2+2x-2)+5.
..
Translating a function right or left is moving the entire curve left or right. In this case moving the curve 4 units to the right means adding -4 units to x like this:f(x)= (x-4)^2+2(x-4)-2
..
Translating the function 5 units up and 4 units to the right look like this:
f(x)=(x-4)^2+2(x-4)-2+5
solving for zeros
f(x)=x^2-8x+16+2x-8-2+5
f(x)=x^2-6x+11
discriminant=b^2-4ac=36-4*1*11<0
Therefore, translated function has no real roots or zeros.

Question 570857: In a circle, if the red line segment is 6 and the blue line segment is 4 how long is the major ellipse
Answer by Alan3354(21608)   (Show Source):
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In a circle, if the red line segment is 6 and the blue line segment is 4 how long is the major ellipse
-----------
What blue line and red line?
Is it a circle, or an ellipse?
There's no "major ellipse"

Question 568252: identify the conic represented by x^2-4y-6x+9=0
Answer by lwsshak3(2927)   (Show Source):
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identify the conic represented by x^2-4y-6x+9=0
complete the square
(x^2-6x+9)-4y+9-9=0
4y=(x-3)^2
y=(1/4)(x-3)^2
This is an equation for a parabola with vertex at (3,0)
Parabola opens upwards with minimum=0.
Standard form of equation for given parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the slope or steepness of the curve.

Question 570123: 1. Find the equation of the parabola with vertex at (3,8) directrix Parabola

2. Find the equation of the parabola with vertex at (5,-1), focus (3,-1)
With the way please
Answer by lwsshak3(2927)   (Show Source):
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1. Find the equation of the parabola with vertex at (3,8) directrix Parabola
(insufficient information to solve)
2. Find the equation of the parabola with vertex at (5,-1), focus (3,-1)
With the way please
Standard form of equation of parabola with given data: (y-k)^2=-4p(x-h), with (h,k) being the (x,y) coordinates of the vertex. Given parabola opens leftward with axis of symmetry, y=-1.
p=distance from the vertex to focus on the axis of symmetry=2
4p=8
Equation of given parabola: (y+1)^2=-8(x-5)

Question 569404: Write an equation of an ellipse given the vertices (-2,4) & (-2,-2) and co-vertices (-3,1) & (-1,1).
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
Write an equation of an ellipse given the vertices (-2,4) & (-2,-2) and co-vertices (-3,1) & (-1,1).
**
Standard form of equation for an ellipse with vertical major axis:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
..
For given ellipse:
center: (-2,1)
length of vertical major axis=6=2a
a=3
a^2=9
length of minor axis or co-vertices=2=2b
b=1
b^2=1
Equation of given ellipse:
(x+2)^2/1+(y-1)^2/9=1

Question 569409: 3(y-3)=(x-6)^2
What steps do I need to do in order to make that equation to y=a(x-h)^2 +k form ?
Answer by lwsshak3(2927)   (Show Source):
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3(y-3)=(x-6)^2
What steps do I need to do in order to make that equation to y=a(x-h)^2 +k form
**
3(y-3)=(x-6)^2
3y-9=x^2-12x+36
3y=x^2-12x+45
divide by 3
y=(1/3)x^2-4x+15
complete the square
y=(1/3)(x^2-4x+4)+15-4/3
y=(1/3)(x-2)^2+41/3
a=1/3, h=2, k=41/3

Question 569682: How do you graph
(x-2)^ 2+(y+ 7)^ 2= 81
Answer by lwsshak3(2927)   (Show Source):
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How do you graph (x-2)^ 2+(y+ 7)^ 2= 81
Standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k) being the (x,y) coordinates of the center, and r=radius
For given equation, (x-2)^ 2+(y+ 7)^ 2= 81
This is a circle with center at (2,-7) and radius of 9

Question 569463: y^2/36-x^2/4=1. find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola and graph
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
y^2/36-x^2/4=1. find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola and graph.
**
Standard form of equation for a hyperbola with vertical transverse axis:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
..
For given hyperbola, y^2/36-x^2/4=1.
center: (0,0)
a^2=36
a=√36=6
vertices: (0,0±a)=(0,±6)=(0,-6) and (0,6)
..
b^2=4
b=2
c^2=a^2+b^2=36+4=40
c=√40≈6.32
Foci: (0,0±c)=(0,±√40)=(0,-6.32) and (0,6.32)
..
Asymptotes:
slope of asymptotes: ±a/b=±6/2=±3
equation of asymptotes: y=3x and y=-3x
see graph below:
y=(36+9x^2)^.5

Question 569207: 27x^2+9y^2=81
Answer by Alan3354(21608)   (Show Source):
You can put this solution on YOUR website!
27x^2+9y^2=81
--------------
Do you have a question?

Question 568943: How would I find the equation of an ellipse if I am give the foci(5,0) and (-5,0) and the length of major axis is 12
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
How would I find the equation of an ellipse if I am give the foci(5,0) and (-5,0) and the length of major axis is 12
**
Standard form of equation for ellipse with horizontal major axis:
(x-h)^2/a^2+(y-k)^2/b^2=1,a>b, (h,k) being the (x,y) coordinates of the center.
..
center: (0,0)
length of major axis=12=2a
a=6
a^2=36
..
c=5
c^2=25
..
c^2=a^2-b^2
b^2=a^2-c^2=36-25=11
..
Equation of given ellipse:
x^2/36+y^2/11=1

Question 568176: find the equation of the ellipse satisfying the given conditions:
canter at (0,0); foci at (0,4) and (0,-4); vertices at (0,7) and (0,-7); and major axis along the y-axis
Answer by Edwin McCravy(6938)   (Show Source):
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find the equation of the ellipse satisfying the given conditions:
canter at (0,0); foci at (0,4) and (0,-4); vertices at (0,7) and (0,-7); and major axis along the y-axis.

It has equation + = 1 and a = 7, so we have a² = 49 + = 1 Here are a bunch of ellipses with center (0,0) and vertices at (0,7) and (0,-7). To find out which it is, we must calculate "b", the semi-minor axis: The relationship between a, b, and c in all ellipses is c² = a² - b² where c is the distance from the center to either focus. The foci are (0,±4) and the vertices are (0,±7) So c = 3 c² = a² - b² 4² = 7² - b² 16 = 49 - b² -33 = -b² 33 = b² = b So the equation is: + = 1 + = 1 And the graph is: Edwin

Question 567809: x^2+y^2=36 what is the center of the ellipse
Answer by Earlsdon(6103)   (Show Source):
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The equation you present, , is that of a circle, which is an ellipes whose major and minor axes are equal.
The center of the circle is at (0, 0) and its radius is 6.

Question 566563: X^2-10x+37+4y^2+16y=0
I need the following:
Center
foci
Major axis
Minor Axis
I get stuck due to -8 issue

I a stuck
Answer by solver91311(12126)   (Show Source):
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Your -8 issue is a self-inflicted wound. Properly done, -8 never appears anywhere in this problem. Also, take care how you type your expressions in the future: X and x are NOT the same thing, so technically you have a three variable equation. Just sayin'

First complete the squares:

Constants on the right!

Divide the 1st order term coefficient by 2, square the result, add the new constant to both sides: -10 divided by 2 is -5, -5 squared is 25.

Factor a 4 out of both terms:

4 divided by 2 is 2. 2 squared is 4. 4 inside the parentheses, 16 outside.

Factor the two perfect square trinomials and collect terms in the RHS:

Divide through by the RHS constant so that the RHS becomes 1:

Compare to the equation of an ellipse centered at , axes parallel to the coordinate axes, major axis measuring , minor axis measuring , and foci at a distance of from the center on the major axis.

So your center is at . Foci at and . Major axis measures 4. Minor axis measures 2.

John

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

Question 564980: The problem is 9y^2+18y=25x^2+216
I have tried over and over to solve this and I keep getting odd numbers. I do not know how to get it into standard form correctly or get a, b, or c to graph it. Help!!
Thank you,
Kaite
Answer by issacodegard(60)   (Show Source):
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Hi, the equation of a hyperbola is

So, consider the left hand side of the equation given in the problem. We can complete the square by adding 9 to both sides. This gives us:

Rearranging, and letting X=x, Y=y+1, we get:

This is the equation of a hyperbola for say z and w:

Shifted one unit down (we would need to take w=y+1 to get the equation given in X,Y)

Question 563654: This hyperbola is centered at (21, -14), and the length of its transverse axis is 24. Which of the equations below could be its equation?
A. (y-14)^2/15^2-(x+21)^2/12^2=1
b. (x-21)^2/15^2-(y+14)^2/12^2=1
c. (x-21)^2/12^2-(y-14)^2/15^2=1
d. (x-21)^2/12^2-(y+14)^2/15^2=1

Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
This hyperbola is centered at (21, -14), and the length of its transverse axis is 24. Which of the equations below could be its equation?
A. (y-14)^2/15^2-(x+21)^2/12^2=1
b. (x-21)^2/15^2-(y+14)^2/12^2=1
c. (x-21)^2/12^2-(y-14)^2/15^2=1
d. (x-21)^2/12^2-(y+14)^2/15^2=1
**
Standard form of equation for hyperbola with horizontal transverse axis:
(x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
For given hyperbola:
center: (21,-14)
Transverse axis=24=2a
a=12
a^2=12^2
..
By process of elimination:
A) does not have correct center coordinates
b) has a^2=12^2 listed under the y-term which means the transverse axis is vertical and the y-term should be listed first before the x-term, but it is not.
c) also does not have correct center coordinates
d) it has the correct center coordinates and a^2=12^2 is listed under the x-term which means the transverse axis is horizontal and the x-term should be listed first as it is. So, this is the best answer.

Question 563737: graph(-f(x))
Answer by fcabanski(385)   (Show Source):
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This is an incomplete problem.

If you need help understanding math so you can solve these problems yourself, then one on one online tutoring is the answer ($30/hr). If you need faster solutions with guaranteed detailed answers, then go with personal problem solving ($3.50-$5.50 per problem). Contact me at fcabanski@hotmail.com

Question 563739: determine with and without graphing whether the function is even, odd, or neither.
(x)/(x^2-3)
x^6-2x^4-1
Answer by richard1234(4802)   (Show Source):
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Suppose f(x) equals the given function. Then

--> odd function

--------

--> even function

Question 563299: where are the foci for an ellipse with vertices at (-5,0) and (5,0) and co-vertices at (0,3) and (0,-3) located?
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
where are the foci for an ellipse with vertices at (-5,0) and (5,0) and co-vertices at (0,3) and (0,-3) located?
**
Standard form of equation for ellipse with horizontal major axis:
(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
..
Equation of given ellipse: x^2/25+y^2/9=1
For given ellipse:
Center: (0,0)
a=5
a^2=25
b=3
b^2=9
c^2=a^2-b^2=25-9=16
c=√16=4
Foci: (±c,0)=(±4,0)=(-4,0) and (4,0)

Question 563407: how do you find the (h,k) of the equation x^2+y^2+16x-4y+59=0
Answer by richard1234(4802)   (Show Source):
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Assuming (h,k) is the center of the circle, complete the square:

The coordinates of the center are (-8,2).

Question 561581: what is the length of the transverse axis of the hyperbola whose equation is (x-7)^2/9^2-(y-4)^2/5^2=1
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
what is the length of the transverse axis of the hyperbola whose equation is
(x-7)^2/9^2-(y-4)^2/5^2=1
This is an equation for a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
..
a^2=9^2
a=9
length of horizontal transverse axis=2a=18

Question 563053: Find center, vertices, and foci of the ellipse: (x+1)^2/16+(y-3)^2/25=1
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
Find center, vertices, and foci of the ellipse: (x+1)^2/16+(y-3)^2/25=1
**
Standard form of equation for an ellipse with vertical major axis:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, with (h,k) being the (x,y) coordinates of the center.
For given elllipse: (x+1)^2/16+(y-3)^2/25=1
center: (-1,3)
a^2=25
a=√25=5
Vertices: (-1,3±a)=(-1,3±5),=(-1,-2) and (-1, 8)
b^2=16
b=√16=4
Foci:
c^2=a^2-b^2=25-16=9
c=√9=3
Foci: (-1,3±c)=(-1,3±3),=(-1,0) and (-1, 6)

Question 562869: Find the equation of a parabola that has a vertex of (9,13) and a y-intercept of 94. Write answer in general form.
Answer by lwsshak3(2927)   (Show Source):
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Find the equation of a parabola that has a vertex of (9,13) and a y-intercept of 94. Write answer in general form.
**
Standard form of equation for parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
..
Plug in given coordinates of vertex and point (0,94), and solve for A.
94=A(0-9)^2+13
94=81A+13
81=81A
A=1
Equation:
y=(x-9)^2+13

Question 562306: what is the standard form of y^2-2y-4x-31=0
Answer by lwsshak3(2927)   (Show Source):
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what is the standard form of y^2-2y-4x-31=0
**
y^2-2y-4x-31=0
complete the square
(y^2-2y+1)-4x-31-1=0
(y-1)^2-4x-32=0
(y-1)^2-4(x+8)=0
(y-1)^2=4(x+8)
This is a standard form of equation for a parabola with vertex at (-8,1)
Parabola opens rightwards with axis of symmetry at y=1

Question 562473: what is the equation for a hyperbola with a center (0,0) co-vertex (o,-9)and focus (-41,0)
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
what is the equation for a hyperbola with a center (0,0) co-vertex (o,-9)and focus (-41,0)
**
Standard form of equation for a hyperbola with horizontal transverse axis:
(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
For given hyperbola:
center: (0,0)
length of conjugate axis(co-vertex)=18=2b
b=9
b^2=81
c=41
c^2=a^2+b^2
a^2=c^2-b^2=41^2-81=1681-81=1600
a=40
Equation:
x^2/1600-y^2/81=1

Question 562068: I am a freshman in honors algebra 2 and we're working on applying prabolas to real world senarios. The question is " find the demensions and maximum area of a rectangle with a perimeter of 48 inches" all I have now is basic rectangle properties because I spaced off in class. Please help me.
Answer by Edwin McCravy(6938)   (Show Source):
You can put this solution on YOUR website!
I am a freshman in honors algebra 2 and we're working on applying prabolas to real world senarios. The question is " find the dimensions and maximum area of a rectangle with a perimeter of 48 inches" all I have now is basic rectangle properties because I spaced off in class. Please help me.

A = L×W where A = area, L = length and W = wifth P = 2L + 2W where P = perimeter Let the length be x Since the perimeter is 48, we substitute in P = 2L + 2W 48 = 2x + 2W Divide every term by 2 24 = x + W Solve for W 24 - x = W Let the area be y. Substitute in A = L×W y = x(24 - x) That equation has this graph: Since y = the area, the area will be a maximum at the peak point, known as "the vertex". There are three ways to find the vertex. I don't know which way your teacher will require you to use. Method 1 for finding the vertex: Find the two x-intercepts y = x(24 - x) x(24 - x) = 0 x=0, 24-x = 0 -x = -24 x = 24 The x-intercepts are (0,0) and (24,0) The vertex occurs halfway when the value of x is halfway between 0 and 24, or at x=12. Substituting x=12 y = 12(24-12) y = 12(12) y = 144 So the vertex of that parabola is (12,144) That means the area y will have a maximium area of 144 square units when the length is x = 12 inches. That is, when the rectangle is a 12in × 12in square. ----------------------------------------- Method 2 for finding the vertex: y = x(24 - x) Put the equation in the vertex form: y = a(x-h)² + k y = x(24 - x) y = 24x - x² y = -x² + 24x Factor out the coefficient of x², which is -1 y = -1[x² - 24x ] Complete the square by multiplying the coefficient of x by ans squaring: -24() = -12, Squaring (-12)² = 144 Add and subtract 144 inside the parentheses: y = -1[x² - 24x + 144 - 144] Factor the first three terms in the brackets as the square of a binomial y = -1[(x - 12)² - 144] Remove the bracket leving the paretheses intact: y = -1(x - 12)² + 144 Compare to y = a(x-h)² + k Vertex = (h,k) = (12,144) ------------------------------------- Method 3 for finding the vertex: Use the vertex formula: The vertex is (h,k) where h = and k = y = -x² + 24x y = -x² + 24x + 0 a = -1, b = 24, c = 0 h = and k = h = = -12 k = k = -1(12)^2+24(12)+0 k = -(144)+ 288 k = 144 So the vertex is (12,144) Use whichever method for finding the vertex your teacher expects you to use, not necessarily the easiest way. Edwin

Question 561628: Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form. This makes absolutely no since to me at all, and I'm super confused. PLEASE HELP ME!
Answer by stanbon(48568)   (Show Source):
You can put this solution on YOUR website!
Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form.
----
Complete the square:
y = x^2+3x + 9 -9-28
---
y = (x+3)^2-37
----
Vertex: (-3,-37)
------------------------
y-intercept:
Use the form y = x^2+3x-28
Let x = 0, then y = -28
-------------------------------
x-intercepts:
Solve x^2-3x-28 = 0
Factor:
(x-7)(x+4) =
x = 7 ; x = -4
--------------------
Cheers,
Stan H.
===============

Question 561358: 162x^4-2y^4

how do i factor this? with work
Answer by jim_thompson5910(21685)   (Show Source):
You can put this solution on YOUR website!
162x^4-2y^4

2(81x^4-y^4)

2((9x^2)^2-(y^2)^2)

2(9x^2-y^2)(9x^2+y^2)

2((3x)^2-y^2)(9x^2+y^2)

2(3x-y)(3x+y)(9x^2+y^2)

So 162x^4-2y^4 completely factors to 2(3x-y)(3x+y)(9x^2+y^2)

In other words, 162x^4-2y^4 = 2(3x-y)(3x+y)(9x^2+y^2)
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Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim
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Question 560429: identify the vertex, focus and directrix. Then sketch the graph
x= -1/4y^2+1/2y-9/4
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
identify the vertex, focus and directrix. Then sketch the graph
x= -1/4y^2+1/2y-9/4
This is an equation of a parabola with standard form: (y-k)^2=4p(x-h), with (h,k) being the (x,y) coordinates of the vertex.
complete the square
x=-1/4(y^2-2y+1)-9/4+1/4
x=-1/4(y-1)^2-2
(x+2)=-1/4(y-1)^2
(y-1)^2=-4(x+2)
vertex: (-2,1)
Axis of symmetry: y=1, parabola opens leftwards
4p=4
p=1
Focus: (-3,1) (p-distance from vertex on axis of symmetry)
Directrix: x=-1 (perpendicular to axis of symmetry and p-distance from vertex on axis of symmetry.
see graph below:
y=±(-4x-8)^.5+1

Question 560874: Find the vertex, the line of symmetry, the maximum or minimum calue of the quadratic function, and graph the function.
f(x)=-2x^2+2x+2
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function.
f(x)=-2x^2+2x+2
complete the square
y=-2(x^2-x+1/4)+2+1/2
y=-2(x-1/2)^2+5/2
This is an equation for a parabola of the standard form: y=-A(x-h)^+k, (hk) being the (x,y) coordinates of the vertex. Negative lead coefficient means parabola opens downwards.
For given function:
vertex: (1/2,5/2)
axis (line) of symmetry: x=1/2
maximum value: 5/2
see graph below:

Question 559257: how would you solve and graph an ellipses with vertices:(4,3) and (4,9) and the focus at (4,8)?

Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
how would you solve and graph an ellipses with vertices:(4,3) and (4,9) and the focus at (4,8)?
**
Given data show this is an ellipse with a vertical major axis. Its equation of standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
..
For given ellipse:
center:(4,6)
length of vertical major axis=12=2a
a=6
a^2=36
c=4
c^2=16
..
c^2=a^2-b^2
b^2=a^2-c^2=36-16=20
b=√20≈4.47
..
Equation of given ellipse:
(x-4)^2/20+(y-6)^2/36=1
see graph below:
y=±(36-(36/20)(x-4)^2)^.5+6

Question 558656: What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
What is the equation of the ellipse with foci (0, 3), (0, -3) and co-vertices (1, 0), (-1, 0)?
Given data shows this is an ellipse with vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
..
For given ellipse:
center: (0,0)
c=3 (from foci)
c^2=9
b=1 (from co-vertices)
b^2=1
c^2=a^2-b^2
a^2=c^2+b^2=9+1=10
a=√10≈3.16
Equation of given ellipse:
x^2/1+y^2/10=1

Question 559127: how do you solve (x+2)^2/9- (y+8)^2/4= 1
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
how do you solve (x+2)^2/9- (y+8)^2/4= 1
This is an equation for a hyperbola with horizontal transverse axis of the standard form:
(x-h)^/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
..
For given equation:
center: (-2,-8)
a^2=9
a=√9=3
b^2=4
b=√4=2
c^2=a^2+b^2=9+4=13
c=√13≈3.6
Vertices: (-2±a,-8)=(-2±3,-8)=(-5,-8) and (1,-8)
Foci: (-2±c,-8)=(-2±√13,-8)=(-2±3.6,-8)=(-5.6,-8) and (1.6,-8)

Question 559049: What is the standard form for a parabola that passes through the points (2,3) and has a vertex of (0, -2)
Answer by lwsshak3(2927)   (Show Source):
You can put this solution on YOUR website!
What is the standard form for a parabola that passes through the points (2,3) and has a vertex of (0, -2)
**
Standard form of equation for a parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
For given parabola:
y=A(x-h)^+k
3=A(2-0)^2-2
3=4A-2
4A=5
A=5/4
Equation:
y=(5/4)x^2-2

Question 558739: What is the vertex, focus, axis of symmetry, and directiv of the following equation, (x-2)² = y+3 and how did you get those answers?
Found 2 solutions by Edwin McCravy, KMST:
Answer by Edwin McCravy(6938)   (Show Source):
You can put this solution on YOUR website!
The standard form of a parabola whose axis is symmetry is vertical is (x - h)² = 4p(y - k) [Some books use "a" or "c" instead of "p"] Where (h,k) is the vertex. |p| is the diatance from the vertex to the focus (which is a point inside the parabola on its axis of symmetry), and also to the directrix, which is a line outside the parabola perpendicular to its line of symmetry. If p is positive the parabola opens upward, and if p is negative it opens downward. We compare your equation to that one: (x - 2)² = y + 3 To get it looking like (x - h)² = 4p(y - k) we put parentheses around the right side and a 1 infront (x - 2)² = 1(y + 3) So we see that h = 2, k = -3, and 4p = 1 which makes p = So the vertex is (h,k) = (2,-3). We plot the vertex (2,-3), and draw a green axis of symmetry through it. That green axis of symmetry goes through x = 2, so that's its equation. The vertex is a point p or of a unit above the vertex. It is on the axis of symmetry so it's x-coordinate is the same as the x-coordinate of the vertex, which is 2, but its y-coordinate is of a unit more, so we add to the y-coordinate of the vertex: -3+ = + = , So the focus has the coordinates (2,) We draw the focus: The directrix is a horizontal line p or of a unit below the vertex We draw it in blue: Since the line is unit below the vertex, we subtract from its y-coordinate -3- = - = , so the equation of the directrix is y = We draw two adjacent squares, with a common side from the directrix to the focus, like this: and sketch in the parabola through the upper corners of those squares and through the vertex: Edwin

Answer by KMST(600)   (Show Source):
You can put this solution on YOUR website!
The equation is the equation of a parabola in vertex form.
VERTEX
The coordinates of the vertex are shown in the equation subtracted from the and the .
The vertex is (2,-3).
The vertex has <---> and <---> , and .
It is a minimum, because, for any value of other than , , making <---> .
AXIS OF SYMMETRY
For all points other than the vertex, the same value of happens for two different values of , at equal distances to the left and right of the line <---> . That line is the axis of symmetry.
FOCUS AND DIRECTRIX
The focus is the point (2,-3+c) above the vertex/minimum that the parabola "wraps" around. The directrix is the line at the same distance on the other side of the vertex.
Your book will tell you that the coefficient of in the equation equals , so in this case -->
The focus has . It is the point (2,-11/4).
The directrix is the line --> .

Question 558697: (y+1)^(2)=-8(x-3)
vertex: ?
focus point: ?
equation of the axis of symmetry: ?
equation of directrix: ?
2 random points on equation: ?
IN NEED OF HELP!!! PLEEEEASE HELP ME ON THIS PROBLEM, I AM REALLY STUCK ON IT AND TRY TO GET BACK WITH ANSWERS ASAP! THANK YOU ONCE AGAIN!
Answer by KMST(600)   (Show Source):
You can put this solution on YOUR website!
The equation is the equation of a parabola in vertex form. In that form, the equation shows you the coordinates of the vertex. They are subtracted from the and the ). The location of the directrix and the focus point can be calculated from the coefficient in the equation.
Your book tells you all about it, but you can figure it all out without the book, and without memorizing any formulas.
VERTEX AND AXIS OF SYMMETRY
What happens when <---> ?
---> --->
The fact that both parenteses are zero tells you that it is a very special point.
For any other value of , ---> --->
So, , no matter what value y takes. It's a maximum value for x. Point (3,-1) is the vertex, and is the axis of symmetry.
TWO RANDOM POINTS
Let's try <--->
---> --->
Let's try <--->
---> --->
DIRECTRIX AND FOCUS POINT
The directrix will be and the focus will be at (3-c,-1)
The points on the parabola for will be at a distance
, measured vertically, above and below the focus point.
They will also be at a distance measured horizontally from the directrix.
By the definition of parabola, those distances are the same, so
--> -->
On the other hand, from the equation given, , we know that for
-->
Substituting that expression for in we get
--> --> is the solution we need.
The directrix is -->
Thee focus point has , so it's the point (1,-1).