Finding an equation for an ellipse using a set of points? Also need help with hyperbolas...?

Question

Hi I'm having a lot of trouble with ellipses and hyperbolas, one of my questions asks "Find the equation for the ellipse with center at the origin, one vertex at (17, 0), and a focus at (8, 0). " and I'm not quite sure how to do it. Also I was wondering, how would I find the x-intercepts for a hyperbola given the equation (x^2/4) + (y^2/9) = 1?

Also, how would I find the foci of an ellipse given the equation 9x^2 + 25y^2 = 225 or the foci of an ellipse given the equation (x^2/4) + (y^2/9) = 1?

Sorry for asking so many questions, just trying to make sure I have everything covered. Any help is appreciated!

Answer 1

Hi,

Find the equation for the ellipse with center at the origin, one vertex at (17, 0), and a focus at (8, 0).

An equation with its center at the origin has the equation:

x²...........y²
-----..+..----- = 1
a²...........b²

Since the distance from the center to a vertex on the major axis is "a", then "a" = 17. The distance from the center to a focus is the value of "c", so "c" equals 8. To find "b", use the formula b² + c² = a².

In this case, b² + 8² = 17² or b² + 64 = 289
b² = 225, so b = 15.

So the equation of the ellipse is:

x²...........y²
-----..+..----- = 1
289.....225


How would I find the x-intercepts for a hyperbola given the equation (x^2/4) + (y^2/9) = 1?

The center is at (0,0). The equation you gave was actually an ellipse because of the "+" in the middle. Its x intercepts would be at (-2,0) and (2,0).

If you wanted the equation (x^2/4) - (y^2/9) = 1 , which would be a hyperbola, would still have x intercepts at (-2,0) and (2,0).



How would I find the foci of an ellipse given the equation
9x^2 + 25y^2 = 225

First, divide everything by 225.


x²........y²
----.+.----- = 1
25.......9

An equation with its center at the origin has the equation:

x²...........y²
-----..+..----- = 1
a²...........b²

So, in our equation, the center is at (0,0).
The vertices are at (-5,0), (5,0), (0,3), and (0,-3).

To find "c", use the formula b² + c² = a².

In this case, 3² + c² = 5²
9 + c² = 25.
c² = 16
c = 4
So foci are at (-4,0) and (4,0).

How would I find the foci of an ellipse given the equation (x^2/4) + (y^2/9) = 1

x².......y²
----.+.--- = 1
4.......9

Since 9 > 4, the major axis of the ellipse goes in the y direction, so the foci will be above and below the center.

The center is at (0,0).
The vertices are at (0,3), (0,-3), (-2,0), and (2,0).

To find "c", use the formula b² + c² = a².

In this case, 2² + c² = 3²
4 + c² = 9.
c² = 5
c = √5
So foci are at (0,√5) and (0,-√5).



I hope that helps!! :-)

Answer 2

For an ellipse with equation x²/a² + y²/b² = 1, x intercepts are ±a, y intercepts are ±b, foci are ±c, where a²-b² = c².

So if center at origin and a vertex at (17,0) and focus at (8,0), a² = 17² = 289, c² = 8² = 64, so b² = 289 - 64 = 225, so b = 15, and the equation is

x²/289 + y²/225 = 1.

If 9x² + 25y² = 225, then
x²/25 + y²/9 = 1, and c² = 25 - 9 = 16, so c = 4, and the foci are (4,0) and (-4,0).

For x²/4 + y²/9 = 1, 9-4 = 5, so foci are (0,√5) and (0,-√5), and that's an ellipse, not a hyperbola. For hyperbolae (that's the Latin plural), one of the terms is negative. If you had x²/4 - y²/9 = 1, then c² = 4 + 9 = 13, and foci are (√13,0) and (-√13,0).

Answer 3

ellispe: (X-H)^2/A^2 + (Y-K)^2/B^2 = 1 where C^2 = A^2-B^2
center: (h,K)
foci: (h + or - c, k)
major axis verticies: (H + or - A, K)
minor axis verticies: ( H, K + or - B)

H=0
K=0
A= 17
B=15
C=8
(X-0)^2/17^2+ (Y-0)^2/15^2 = 1
final equation: X^2/284 + Y^2/225 = 1

Work:

focus (8, 0) with foci being (h + or - c, k)
We know that H and K are 0.
H + C = 8
0+C=8
C=8

vertex at (17, 0) this is a major axis vertex because we're adding to H. a maj. axis vertex is: (H + or - A, K)
H + A = 17
0+A=17
A=17

in an elipse, C^2=A^2-B^2
8^2=17^2-B^2
64=289-B^2
-225=-B^2
225=B^2
15=B
----------------------------------
for the hyperbola, you might want to graph it and just do it that way.

9x^2 + 25y^2 = 225
make it an equation we can read better.
9x^2 + 25y^2 = 225
Divide EVERYTHING by 225
9x^2/225 + 25y^2/225 = 225/225
x^2/75 + y^2/9 = 1

remember that foci are (H+ or - C, K)
C^2= 75-9
=66
=sq.rt.66
so (sq.rt.66, 0) and (-sq.rt.66, 0)

Do the same thing for the other equation. Use (H + or - C, K) and C^2=A^2-B^2

Answer 4

1) When centered at the origin, one vertex tell you the value of A ( here 17) and the other is at (-17,0) Horiz. Major Axis

Focucs is C (here c = 8 ) and c^2 = a^2 - b^2

so 64 = 289 - b^2 and B = 15 hence (0,15) (0,-15)

x^2 / 189 + y^2 / 225 = 1



2) To find intercepts, replace y with zero and solve for x


3) Foci are found only by the formula c^2 = a^2 - b^2

A and B are the denominators of the fractions when in the form x^2/a^2 + y^2 / b^2 = 1