Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
\(y =\pm \dfrac 56 x\)

Answer 1

I think it is \(\dfrac{x^2}{100} - \dfrac{y^2}{144} = 1\)

Answer 2

@ganeshie8

Answer 3

give vertices are \(0,~\pm 10\)

theat means the hyperbola opens up and down, yes ?
(Vertical hyperbola)

Answer 4

Yes

Answer 5

oh, so it would be \(\dfrac{y^2}{100} - \dfrac{x^2}{144} = 1\)

Answer 6

Yep!
http://www.wolframalpha.com/input/?i=asymptotes+y%5E2%2F100-x%5E2%2F144%3D1

Answer 7

Thanks! :)

Answer 8

np:)