Question

What is the equation, in standard form, of a vertical hyperbola with asymptotes at y + 8 = ±2/9(x – 8)?

Answer 1

Vertical hyperbola has asymptotes:
\[y=\pm \frac{ a }{ b }(x-h)+k\]
You have:
\[y = ±\frac{ 2 }{ 9 }(x – 8) - 8\]

and the standard form of a vert. hyperbola:
\[\frac{ (y-k)^2 }{ a^2 }-\frac{ (x-h)^2 }{ b^2 } = 1\]

Answer 2

what is the y?

Answer 3

I'm not sure what you mean...

Answer 4

y is a variable like x

Answer 5

yeah, but in y= ±2/9(x – 8)-8, isthere a value for y?

Answer 6

to use to convert it to standard form?

Answer 7

No, there's no need for y or x values. If you look at my first reply, compare the first two formulas - they give you a, b, h and k.

Answer 8

Once you find a, b, h, k by comparing the first two, insert them into the second equation for the standard form.

Answer 9

k= -8
h=8
a=2

Answer 10

Correct, and b?

Answer 11

9

Answer 12

Yep! Now just enter them into:
\[\frac{ (y-k)^2 }{ a^2 }-\frac{ (x-h)^2 }{ b^2 } = 1\]

Answer 13

(y+8)^2/4- (x+8)^2/81=1

Answer 14

The (x+8) should be (x-8), since h = 8

Answer 15

Change that and it all looks correct.