Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola:
x^2+9y^2+36y-72=0

Answer 1

Put the 72 on the right side.

Answer 2

Then write
\[9y^2+36y=9(y^2+4y+\]

Answer 3

And complete the square.

Answer 4

so ignore the x^2?

Answer 5

Only for a moment.

Answer 6

Just until we get that y slop taken care of.

Answer 7

okay.

Answer 8

so is it x^2+9(y+2)^2=108?

Answer 9

Wow!! I am impressed. Yes. That's it.
Now divide every term by 108

Answer 10

Oh. Something is wrong. This is not a hyperbola. Is there a typo somewhere?

Answer 11

my teacher said it is a hyperbola so.

Answer 12

Are you sure there is not a sign error? In hyperbolas, either x2 or y^2 is negative.

Answer 13

If they are both positive, it is either an ellipse or a circle.

Answer 14

whoops there is a sign error its supposed to be a -9y^2. sorry, my bad.

Answer 15

So we need to rethink that y part because now it is: -9(y^2-4y

Answer 16

yeah you're right. so it would be like x^2-9(y-2)^2=36?

Answer 17

Yes. That's it. Now divide each term by 36

Answer 18

okay so it'd be \[\frac{ x^2 }{ (6)^2 }- \frac{ (y-2)^2 }{ (2)^2 }=1 \]

Answer 19

Yes. So what is the center?

Answer 20

(0,2)

Answer 21

And what is the value of a?

Answer 22

is it 0?

Answer 23

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]

Answer 24

Your equation matches that one. h=0, k = 2, what is a?

Answer 25

it would equal 6.

Answer 26

Since x^2 is positive, this hyperbola opens to the right and to the left. And the vertices are 6 units to the right and the left of the center.

Answer 27

So what are the vertices?

Answer 28

okay. so it'd be (-6,2) and (6,2)

Answer 29

Yes. Now we should probably put this information on a graph.

Answer 30

Did your teacher teach you to draw the rectangle based on the a and b values?

Answer 31

nope not really. he just showed us the math to get the information. he sorta skipped over it.

Answer 32

a=6 and b = 2. So if we go a units right and left of the center, and b units up and down from the center we can draw a rectangle based on those points.