Precalc help please?

Write the equation of the hyperbola whose center is at the origin and has a vertical transverse axis. The equations of the asymptotes are 6x+2y=0 and 6x-2y=0.

So, I keep getting (y^2/9)-(x^2/4)=1, but that can't be it because the asymptotes of this are +/-3/2x and they're supposed to be +/-3x?

Question

Precalc help please?

Write the equation of the hyperbola whose center is at the origin and has a vertical transverse axis. The equations of the asymptotes are 6x+2y=0 and 6x-2y=0.

So, I keep getting (y^2/9)-(x^2/4)=1, but that can't be it because the asymptotes of this are +/-3/2x and they're supposed to be +/-3x?

anonymous · Answer

@ganeshie8 can you help me with this?

beccaboo333 · Answer

I'm sure he can :3 he's like amazing with math <3

ganeshie8 · Answer

lol no, you're being kind of me becca :) conics are tricky

ganeshie8 · Answer

(y^2/9)-x^2=1

ganeshie8 · Answer

this wud give u the required asymptotes

ganeshie8 · Answer

may i knw how u worked the equation you're having ? :)

anonymous · Answer

Oh okay, could you explain how to get that?

ganeshie8 · Answer

i have just fixed it from the asymptotes, but im sure there must be a standard method

anonymous · Answer

And as for what I did before, since b^2 is c^2-a^2 I think I just got the value of c messed up.

ganeshie8 · Answer

you said you're getting 2 extra, so i chopped off 2^2 from the denominator of x^2 so that asymptotes look correct

ganeshie8 · Answer

do u work it by drawing rectangle  ?

anonymous · Answer

Oh okay, I was basically just guessing and checking before actually :/

anonymous · Answer

I have to go, but thank you for your help even though I still don't fully understand :P

beccaboo333 · Answer

Sorreh I didn't help. But Ganesh is well known for helping in math :3

ganeshie8 · Answer

with the given asymptotes we can oly find the relation between "a" and "b"
$\large \frac{a}{b} =  \pm 3$

ganeshie8 · Answer

to find the exact equation of hyperbola, we will need atleast "one point" on the hyperbola. 

since we're just given asymptotes and not given any point, there will be infinitely many hyperbolas with these asymptotes

ganeshie8 · Answer

for ex, below is another hyperbola having the same asymptotes :-

$\large  \frac{y^2}{18} - \frac{x^2}{2} = 1$

ganeshie8 · Answer

basically, asymptotes are not sufficient to uniquely identify a hyperbola. we need a "point" also

ganeshie8 · Answer

when u get back, let me knw if smthng doesnt make sense :)

anonymous · Answer

Yeah, I think I understand :) So the fact that the center is at (0, 0) doesn't help us since that isn't a point on the actual hyperbola itself?