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

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- anonymous

- katieb

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- Plasmataco

... I'll try but no garentees

- anonymous

No problem! Atleast if you can try or get someone to help you too

- Plasmataco

It's center is at the origin so...

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## More answers

- Plasmataco

Ok... U listening?

- Plasmataco

What u got to do is put the x in the coordinates of the vertices in for the asymptote equation.

- Plasmataco

So plug 10 in to get y=50/6

- anonymous

is this vertical or horiztonal though

- Plasmataco

.

- Plasmataco

Hang on.

- Plasmataco

Because the center is (0,0), it makes thing a lot easier

- anonymous

Oh okay

- Plasmataco

The equation is ((X-g)/a)-((y-f)/b)=1

- anonymous

okay yeah I have that

- Plasmataco

(G,f) is the center of the hyperbola in which case (0,0) so we can cancel those out to get (X squared)/a-(y squared)/b

- Plasmataco

My first equation is wrong. I forgot the power 2 after(x-g) and (y-f) sry 😓

- anonymous

okay yeah I was gonna correct that

- Plasmataco

A is the x part of the vertice so 10. B is what we solved earlier with the asymptote.

- anonymous

wait what is b?

- Plasmataco

So we get (x squared)/10-(y squared times 6)/50=1

- Plasmataco

The 6th comment

- Plasmataco

No wait the answer is (x/10) squared-(6y/50) squared=1

- Plasmataco

Forgot another pair of squares at the denominator

- Plasmataco

Here's the link I used

- Plasmataco

http://www.endmemo.com/geometry/hyperbola.php

- Plasmataco

@PHUNISH ?

- Plasmataco

Ok well. I gotta go sry

- Plasmataco

I hoped I helped

- anonymous

Yeah sorry I been trying to figure this out with my notes but you helped so much more thank you so much

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