# Thread:

1. ## Equation of Ellipse in Standard Form

I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2.

I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1.

This is a horizontal elipse with center (0,0). The length of the minor axis is 4 units so I think c = 4 and c^2 = 16. I don't know where to go to find a^2 and b^2.

Can you help? Thanks.

Joanie

2. Originally Posted by Joanie
I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2.

I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1.
In any ellipse, the distance between the center of the ellipse and either focus is $c=\sqrt{a^2-b^2}.$ You know $b$ (it is half the length of the minor axis), and you know the distance between the center and the foci. So you can solve for $a^2.$

3. ## Response to Equation of an Ellipse

Thanks. I will do the problem and have you check my answer.

Joanie

4. Originally Posted by Joanie
I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2.

I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1.

This is a horizontal elipse with center (0,0). The length of the minor axis is 4 units so I think c = 4 and c^2 = 16. I don't know where to go to find a^2 and b^2.

Can you help? Thanks.

Joanie
Hi Joanie,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The two foci are (-c, 0) and (+c, 0).

So c would have to be 8, wouldn't it? Not 4.

The y-intercepts are at (0, b) and (0, -b) so b would have to be 2 and $b^2=4$.

Using $c^2=a^2-b^2$, you can easily solve for $a^2$.

Now just plug and chug.
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