I found the perfect square of this ellipse equation, now what's its center?

Question

I spent FOREVER trying to figure out how to find the perfect square of an ellipse equation, AND I FINALLY FIGURED IT OUT! I was sooo excited. Then I looked at the original question: Find the center of the ellipse with the equation 3x^2+4y^2+18x-32y-5=0

I completed the square, and got (x+3)^2 OVER 32 + (y-4)^2 OVER 24 =1


you can obviously replace "OVER" with a fraction sign, I just had to do that so it didn't look like it was 2/32 or 2/24 or anything like that, so I hope it made sense!

My book says that the standard form of the equation of the ellipse is x^2/b^2+y^2/a^2=1, then it says b^2=a^2-c^2

I was thinking I could make my equation look like that by just finding the square roots of 32 and 24, but I thought about it, and they're not perfect squares, so they come out as really random numbers, so I'm guessing that's not what I'm supposed to do??

How do I find the center of this ellipse?! I've spent SO LONG on this problem it's frustrating me!!!!

THANKS!

EDIT: I think I figured it out! Wouldn't it be (-3,4)?!

EDIT AGAIN: * COMPLETING the square, not perfect square

lol
duh, me.


sorry, it's 3 am here and i'm worn out!

Answer 1

Find the center of the ellipse with the equation
3x² + 4y² + 18x - 32y - 5 = 0.

Complete the squares.

3x² + 4y² + 18x - 32y - 5 = 0
3x² + 18x + 4y² - 32y = 5
3(x² + 6x) + 4(y² - 8y) = 5
3(x² + 6x + 9) + 4(y² - 8y+ 16) = 5 + 3*9 + 4*16
3(x + 3)² + 4(y - 4)² = 96

If all you want to do is find the center you can stop here. The center is:

(h, k) = (-3, 4)

You don't need to finish putting the equation into standard form. You already have the answer.