Originally Posted by

**psu1024** I am having trouble finding the standard form of an ellipse, being given the following equation:

$\displaystyle

36x^2+9y^2+48x-36y+43=0

$

I know that I have to group terms and complete the square to make it fit the standard form of an ellipse:

$\displaystyle

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

$

So I get to this and have been stuck:

$\displaystyle

(36x^2+48x+?)+9(y^2-4y+4)=-34+?+36

$

I don't think I can just factor out the 12, and have $\displaystyle

12(3x^2+4x+?) $, so I tried factoring out the 36:

$\displaystyle

36(x^2+\frac{3}{4}x+\frac{9}{64})

$

in which I got $\displaystyle

36(x+\frac{3}{8})^2 + 9(y-2)^2=\frac{113}{16} $ <---which I got from $\displaystyle (-34+36(\frac{9}{64})+36)

$ ,which doesn't work out at all because I can't divide out what's to the right of the equal sign to get a nice looking equation that equals 1.

I'm completely baffled by this problem, so any help would be appreciated.