## GiggleSquid Need your help again rkfitz053! one year ago one year ago

1. GiggleSquid

trying to write it in standard form. i got -4(x^2-8x) + (y^2-12y)=-36 so far

2. rkfitz053

Start by gathering all like terms so the equation looks like this (you got this one.):$-4x^2-8x+y^2-12y=-36$ Then start by completing the square (take 1/2 of the coefficient on the x term and square that, then add it to the end, make sure you also add it to the right side. And multiply that -4 to the 1 you added) for the x terms and the y terms: $-4(x^2+2x+1)+(y^2-12y+36)=-4$ Now you have two perfect square trinomials and you can factor those and divide all terms by -4 to end up with: $(x+1)^2 - (y-6)^2/4=1$ Can you figure out the centre from there?

3. GiggleSquid

ya its -1,6 right? thanks!

4. rkfitz053

Got it. No problem.

5. rkfitz053

The trickiest part is always remembering to multiply what you add by the factor that you took out before (multiplying -4 to 1 when completing the sq)

6. GiggleSquid

how do i find the vertices? i'm better with ellipses :/

7. GiggleSquid

@rkfitz053

8. rkfitz053

Your vertices are going to be a distance of 'a' away from the center. Remember the std eqn for hyperbolas: $x^2/a^2-y^2/b^2=1$ so you just have to take the sqrt of what the first term is divided by. In your question a^2=1

9. GiggleSquid

the foci?

10. rkfitz053

The foci are almost the same as ellipses too...it's just $c=\sqrt{a^2+b^2}$ They're added now because the foci points are on the outside of the vertices, rather than inside an ellipse.

11. rkfitz053

Vertices are the same as ellipses and foci you add instead of subtract.

12. rkfitz053

Got it?

13. GiggleSquid

got it

14. rkfitz053

good stuff