Can someone help walk me through this problem? I'd really appreciate your help.

Find an equation in standard form for the hyperbola with vertices at (0, -10) and asymptotes at y= +-(5/4)x

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- Jan 18th 2013, 11:07 AMItelFind an equation
Can someone help walk me through this problem? I'd really appreciate your help.

Find an equation in standard form for the hyperbola with vertices at (0, -10) and asymptotes at y= +-(5/4)x - Jan 18th 2013, 11:39 AMMarkFLRe: Find an equation
If the transverse axis is on the

*y*-axis, then the form of the hyperbola is:

$\displaystyle \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ where $\displaystyle 0<a,b$.

Since the vertices are (presumably) at $\displaystyle (0,\pm10)$, then what is the value of*a*?

The asymptotes will be at $\displaystyle y=\pm\frac{a}{b}x$, so from the given asymptotes, we know:

$\displaystyle \frac{a}{b}=\frac{5}{4}$

$\displaystyle b=\frac{4}{5}a$

Use the value you found for*a*to find*b*.

and so what is the equation of the hyperbola? - Jan 18th 2013, 12:29 PMItelRe: Find an equation
Would a be -10?

Which would make b 8? - Jan 18th 2013, 12:35 PMMarkFLRe: Find an equation
*a*and*b*are both positive, so $\displaystyle a=10$ and yes, then $\displaystyle b=8$. - Jan 18th 2013, 12:41 PMItelRe: Find an equation
So the equation would just be

y^2/100 - x^2/64 = 1? - Jan 18th 2013, 12:47 PMMarkFLRe: Find an equation
Yes, although the standard form is:

$\displaystyle \frac{y^2}{10^2}-\frac{x^2}{8^2}=1$ - Jan 18th 2013, 01:02 PMItelRe: Find an equation
Thank you so much for your help. I really appreciate it.