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?

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- Mar 24th 2010, 10:48 AMiluvmathbutitshardAm I right?
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? - Mar 24th 2010, 11:04 AMharish21
- Mar 24th 2010, 11:44 AMiluvmathbutitshard
So

x^2 / 15 - y^2 / 16 = 1?

Thanks for answering - Mar 24th 2010, 11:49 AMharish21
- Mar 24th 2010, 11:50 AMiluvmathbutitshard
woops, sorry. Made a mistake :)

x^2 / 9 - y^2 / 16 = 1

Right?

Thank you! - Mar 24th 2010, 12:03 PMharish21
- Mar 25th 2010, 03:16 AMHallsofIvy
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.