Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ±(5/6) x.

Answer 1

@phi @robz8 ?!?!

Answer 2

we can do this i am sure
first off the center is \((0,0)\) so you know it looks like
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

Answer 3

second, since the vertices are at \((0,\pm10)\) we know \(a=10\) so we get to
\[\frac{x^2}{10^2}-\frac{y^2}{b^2}=1\] so all we need now it \(b\)

Answer 4

a little weird at this point though
you sure it is \(y=\frac{5}{6}x\) right?

Answer 5

\[\frac{b}{a}=\frac{5}{6}\] so \[\frac{b}{10}=\frac{5}{6}\] giving \(b=\frac{50}{6}\) hmm

Answer 6

ok if that is what it is, then that is what it is
\[\frac{x^2}{100}-\frac{36y^2}{2500}=1\] lets check it

Answer 7

http://www.wolframalpha.com/input/?i=hyperbola+ \frac{x^2}{100}-\frac{36y^2}{2500}%3D1

yup looks good to me

Answer 8

sorry about that, thats what i thought. THANK YOU

Answer 9

@satellite73