Question: Find an equation of the hyperbola with its center at the origin

Vertices: (+/- 3, 0);

Foci: (+/- 5, 0)

My work:

a = 3, c = 5

b^2 = c^2 - a^2

25 - 9 = 16

(y^2 / 25) - (x^2 / 16) = 1

I don't think this is right. Any corrections?

Results 1 to 7 of 7

- Mar 24th 2010, 11:48 AM #1

- Joined
- Nov 2009
- Posts
- 87

- Mar 24th 2010, 12:04 PM #2

- Mar 24th 2010, 12:44 PM #3

- Joined
- Nov 2009
- Posts
- 87

- Mar 24th 2010, 12:49 PM #4

- Mar 24th 2010, 12:50 PM #5

- Joined
- Nov 2009
- Posts
- 87

- Mar 24th 2010, 01:03 PM #6

- Mar 25th 2010, 04:16 AM #7

- Joined
- Apr 2005
- Posts
- 20,032
- Thanks
- 3160

You've already had this answered but notice that if y= 0, this equation gives you $\displaystyle -\frac{x^2}{16}= 1$ which is impossible. It does NOT pass through (3, 0) or (-3, 0), the vertices.

For $\displaystyle \frac{x^2}{16}- \frac{y^2}{25}= 1$, if y= 0 you have $\displaystyle \frac{x^2}{16}= 1$ which gives $\displaystyle x= \pm 5$ which is still wrong.