Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/2x

Answer 1

any idea?

Answer 2

vertical right?

Answer 3

don't forget asymptotes y = \(\pm\)b/a

Answer 4

ohh so it is y^2/4 - x^2/9 ?

Answer 5

not yet!! you still have a step to figure out which is the form from

Answer 6

vertical

Answer 7

vertex (0, 6) means the vertical form, right!!

Answer 8

ye, you get the correct answer

Answer 9

yay thanks ^_^

Answer 10

oh, we go backward, not that I am sorry

Answer 11

the vertical form have the asymptote formula is y =\(\pm a/b\) so, a = 3, b =2
the standard form is \[\dfrac{y^2}{9}-\dfrac{x^2}{4}=1\]

Answer 12

*\(y=\pm \dfrac{a}{b}x\) I am so sorry for my restlessness.