Question

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

Answer 1

... I'll try but no garentees

Answer 2

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

Answer 3

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

Answer 4

Ok... U listening?

Answer 5

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

Answer 6

So plug 10 in to get y=50/6

Answer 7

is this vertical or horiztonal though

Answer 8

.

Answer 9

Hang on.

Answer 10

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

Answer 11

Oh okay

Answer 12

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

Answer 13

okay yeah I have that

Answer 14

(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

Answer 15

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

Answer 16

okay yeah I was gonna correct that

Answer 17

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

Answer 18

wait what is b?

Answer 19

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

Answer 20

The 6th comment

Answer 21

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

Answer 22

Forgot another pair of squares at the denominator

Answer 23

Here's the link I used

Answer 24

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

Answer 25

@PHUNISH ?

Answer 26

Ok well. I gotta go sry

Answer 27

I hoped I helped

Answer 28

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