# Standard Equation of an Ellipse

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• Jul 23rd 2009, 09:36 PM
psu1024
Standard Equation of an Ellipse
I am having trouble finding the standard form of an ellipse, being given the following equation:
$
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:
$
\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1
$

So I get to this and have been stuck:

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

I don't think I can just factor out the 12, and have $
12(3x^2+4x+?)$
, so I tried factoring out the 36:

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

in which I got $
36(x+\frac{2}{83})^2 + 9(y-2)^2=18$
<---which I got from $(-34+36(\frac{4}{9})+36)
$

So then I get
$
2(x+\frac{2}{3})^2 + \frac {(y-2)^2}{2} = 1
$

Does this mean that $a=\frac{1}{2}$ ? Is that even possible? And if so, does this mean that the center is $(-\frac{2}{3},2)$ ?

In this case then, the major axis is vertical, correct? So the foci would be $(-\frac{2}{3}, 2\pm\sqrt{1.5})$ and the vertices would be $(-\frac{2}{3},2\pm\sqrt{\frac{1}{2}})$ and $(-\frac{2}{3}\pm\sqrt{2},2)$ ?

Compared to what we've been learning in class (even though we haven't really even talked about it all that much), this all just seems too complicated to be correct, so I was didn't know if there was just something I was doing wrong or overthinking something.

Any help would be appreciated.
• Jul 23rd 2009, 09:51 PM
VonNemo19
Quote:

Originally Posted by psu1024
I am having trouble finding the standard form of an ellipse, being given the following equation:
$
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:
$
\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1
$

So I get to this and have been stuck:

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

I don't think I can just factor out the 12, and have $
12(3x^2+4x+?)$
, so I tried factoring out the 36:

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

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

to complete this square...
$36x^2+48x$

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

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

Understand that the first and second lines are equivalent, the third is just for conceptualization. Do you see that you must add 16 to the right hand side to preserve equality?
• Jul 23rd 2009, 10:07 PM
psu1024
Oh, good catch! Thanks. I think I reversed the 36 and 48 when I was dividing them.
• Jul 23rd 2009, 10:09 PM
VonNemo19
Quote:

Originally Posted by psu1024
Oh, good catch!

I hate it when I miss the small stuff. Best of luck!