Question

help me find the asymptotes.
(x-3)^2 -(y+1)^2=9

Answer 1

Is this the entire problem?

Answer 2

I know how to find the center vertices and foci but not the asymptotes

Answer 3

What is b?

Answer 4

the center is (3, -1) transverse axis is the y-axis and the vertices are (3,2) and (3, -4) the foci are (3, sqroot of 18) and (3, -sqroot of 8).

Answer 5

if I did it right.

Answer 6

The x^2 term is positive so this hyperbola opens to the right and left...the transverse axis is horizontal

Answer 7

so it is transverse on the x-axis not the y.

Answer 8

yes

Answer 9

so it makes the vertices (0, -1) and (6, -1)??

Answer 10

yes

Answer 11

and the foci would be (sqroot 18, -1) and (-sqroot 18, -1)

Answer 12

\[c=\sqrt{a^2+b^2}=\sqrt{9+9}=3\sqrt{2}\]

Answer 13

Foci are c units to the right and c units to the left of the center.

Answer 14

yes so i would just replace the sqroot 18 with 3sqroot2

Answer 15

yes. and then add it to the x coordinate of the center and subtract it from the x coordinate of the center.

Answer 16

okay so it would be (3-3sqroot2, -1) and (3+3sqroot2, -1)

Answer 17

So the foci are:
\[(3+3\sqrt{2},-1); (3-3\sqrt{2},-1)\]

Answer 18

okay then how about the asymtotes

Answer 19

The asymptotes are lines. They pass through the center. Their slopes are + - b/a

Answer 20

so wouldnt it just be 1

Answer 21

since a and b are 9