Question

What is the equation, in standard form, of a horizontal hyperbola with asymptotes at y – 6 = ±6/11(x + 6)?

Answer 1

If a hyperbola has an asymptote of the form \[y -h = \pm b(x-k)/a\] then the hyperbola is of the form\[(x-k)^2/a^2 - (y-h)^2/b^2 = 1\]

Answer 2

But the last part is different than the first form the fractions in a weird place.

Answer 3

Here are the answers if it helps
open parentheses x plus 6 close parentheses squared over 36 end fraction minus open parentheses y minus 6 close parentheses over 121 = 1

open parentheses x minus 6 close parentheses squared over 36 end fraction minus open parentheses y plus 6 close parentheses over 121 = 1

open parentheses x plus 6 close parentheses squared over 121 end fraction minus open parentheses y minus 6 close parentheses over 36 = 1

open parentheses x minus 6 close parentheses squared over 121 end fraction minus open parentheses y plus 6 close parentheses over 36 = 1

Answer 4

do the answer choices help you out?

Answer 5

slope are going to be of the form y/x. Squareing the given slope and sticking the top under the y part; and the bottom under the x parts helps to define those a^2 and b^2 terms

Answer 6

I'm sorry, What?

Answer 7

dominus has a good general setup; all you have to do is fill in the parts.

if anything, the a^2 and b^2 parts can be confusing to fill in

Answer 8

The problem I'm having is that it says ±6/11(x + 6) instead of ±(x-k)/a

Answer 9

Like it's reversed.

Answer 10

the x and y parts are just reused

(x+6)^2 and (y-6)^2

since this thing opens horizontaly; its parallel to the x axis, so x is positive and y is negtaive (subtracted)

(x+6)^2 - (y-6)^2

Answer 11

\[\frac 6{11}=\sqrt{\frac yx}\]
\[\left(\frac 6{11}\right)^2=\frac yx\]

Answer 12

the y parts have 6^2 under them; and the x parts have 11^2 under them

Answer 13

So the answer is d!!!!! Thank you so much could NOT have done this without your help!!!