Question

The hyperbola with center at (5, -6), with vertices at (0, -6) and (10, -6), with (15, 0) a point on the hyperbola.

Answer 1

Have you determined whether this is an upward-downward opening hyperbola or a left-right opening hyperbola?

Answer 2

I tried to graph it but the vertices ended up in the same quadrant

Answer 3

Well, that's fine, because the center is not at the origin. The hyperbola is just translated into the first quadrant. Is it up-down or left-right? Plotting the center will help.

Answer 4

*fourth quadrant

Answer 5

left right

Answer 6

Good. A general equation for a left-right opening parabola centered at the origin with radius \(r\) is given by the following. Here \(a\) is an arbitrary constant.\[\Large x^2-ay^2=r^2\]First off: What's the radius \(r\)? Second, translate it so that it's centered about your point. And third, use the given intersection point to find \(a\).

Answer 7

The general form is
\[
\frac{ (x-5)^2}
{25}-\frac{(y+6)^2}{b^2}=1
\]
Replace x by 15 and yby zerp and solve for b

Answer 8

okay I see what your saying so far :)

Answer 9

you get
\[4-\frac{36}{b^2}=1\\
\frac{36}{b^2}=3\\
b^2 =12
\]

Answer 10

did I have to graph any of this

Answer 11

\[
\frac { (x-5)^2}{25}-\frac{(y+6)^2}{12}
=1
\]
No you do not need to graph it.

Answer 12

okay the answer is (x-5)^2/25-(y+6)^2/12=1
and the center is hk which os -5 and 6