1. ## Am 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?

2. Originally Posted by iluvmathbutitshard
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?
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

3. So
x^2 / 15 - y^2 / 16 = 1?

4. Originally Posted by iluvmathbutitshard
So
x^2 / 15 - y^2 / 16 = 1?

how did you get 15?

${a^2} = 9$

${b^2} = 16$

5. woops, sorry. Made a mistake
x^2 / 9 - y^2 / 16 = 1
Right?
Thank you!

6. Originally Posted by iluvmathbutitshard
x^2 / 9 - y^2 / 16 = 1
Right?
Thank you!
looks good to me..

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

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