# Am I right?

• Mar 24th 2010, 10:48 AM
iluvmathbutitshard
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?
• Mar 24th 2010, 11:04 AM
harish21
Quote:

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?

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
• Mar 24th 2010, 11:44 AM
iluvmathbutitshard
So
x^2 / 15 - y^2 / 16 = 1?

• Mar 24th 2010, 11:49 AM
harish21
Quote:

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

how did you get 15?

$\displaystyle {a^2} = 9$

$\displaystyle {b^2} = 16$

• Mar 24th 2010, 11:50 AM
iluvmathbutitshard
woops, sorry. Made a mistake :)
x^2 / 9 - y^2 / 16 = 1
Right?
Thank you!
• Mar 24th 2010, 12:03 PM
harish21
Quote:

Originally Posted by iluvmathbutitshard
woops, sorry. Made a mistake :)
x^2 / 9 - y^2 / 16 = 1
Right?
Thank you!

looks good to me..
• Mar 25th 2010, 03:16 AM
HallsofIvy
Quote:

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 $\displaystyle -\frac{x^2}{16}= 1$ 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.