Question

What is standard form of this hyperbola? 9y^2 - 16x^2 - 144 = 0

Then I can graph it.

Answer 1

Hyperbolas are of the form\[(x-h)^{2}/a +(y-k)^{2}/b=1\]

Answer 2

How do I convert it to that?

Answer 3

Use intuition, from your equation how can you get a one on the right and at the same time some number under x and y?

Answer 4

Add 144 to the 0?

Answer 5

Great, you're on the right track, continue.

Answer 6

Would I now divide the 9 and 16 by 144?

Answer 7

Yes

Answer 8

Is that the final answer?

9y^2/144 - 16x^2/144 = 1

Answer 9

9 can divide 144, and 16 into 144

Answer 10

Divide 9 by 44?

Answer 11

You want a^2 and b^2 at the bottom, so divide 144 by 9 and so on

Answer 12

9y^2/16 - 16x^2/12 = 1

Answer 13

You're on the right track, you should cancel the 9 and 16 and calculations off a little on b\[(y-0)^{2}/4^{2} -(x-0)^{2}/3^{2}=1\]Good job

Answer 14

Thanks