hyperbola

[tsh44]

Hi i have to graph this hyperbola 9x^2-16y^2-72x-64y-64=0. How would I change this to the (Y^2)/(A^2) - (x^2)-(b^2) =1 form or is there another way?

 

[tkhunny]

Hi i have to graph this hyperbola 9x^2-16y^2-72x-64y-64=0. How would I change this to the (Y^2)/(A^2) - (x^2)-(b^2) =1 form or is there another way?

Completing the Square - Twice.

9x² - 16y² - 72x - 64y - 64 = 0

9x² - 72x - 16y² - 64y = 64

9(x² - 8x) - 16(y² + 4y) = 64

9(x² - 8x + ____) - 16(y² + 4y + ____) = 64 + 9*____ - 16*____

-8/2 = -4
(-4)² = 16

4/2 = 2
2² = 4

9(x² - 8x + 16) - 16(y² + 4y + 4) = 64 + 9*16 - 16*4

9(x - 4)² - 16(y + 2)² = 64 + 144 - 64 = 144

(x - 4)²/16 - (y + 2)²/9 = 1

That's about it. You show us the next one.

 

[Gene]

complete the squares.
9x^2-16y^2-72x-64y-64 =
9(x^2-8x+16)-16(y^2+4y+4) -64 - (9*16 + 4*16) =
9(x-4)^2-16(y+2)^2 - 144 = 0

O.K. TKH beat me to it and gave you more than I am. I would have deleted this but I wanted to say DO NOT mix caps & lc for the same variable. Standard practice is that Y is not the same as y. They are two different variables.

 

[tsh44]

Normally there is onyl x^2 and y^2 on top, but in this case there is
(x-4)^2 & (y+2)^2. Does this affect the graphing. I normally fidn the asymptotes by +/-(b/a) * x. Can I use this in this problem?

 

[tsh44]

Ok never mind I found the center to be (4,-2), the vertices to be (8,-2), (0,-2) and the foci to be (9,-2), (-1,-2). I still dont know how to find the equations of the asymptotes however. Can I still use +/- b/a (x) if it's not centered at the origin?

 

[Gene]

That gives the slope of the asymptotes. That doesn't change when you move the figure. The actual line equation has to go thru the (new) center so that would change. Use the point-slope equation.