# Put a Conic Section into General Form

• Mar 22nd 2010, 12:51 PM
sleigh
Put a Conic Section into General Form
I have to put the ellipse with the equation $\displaystyle ax^2+by^2+cx+dy+e=0$ into standard form, $\displaystyle \frac{(x-h)^2}{a}+\frac{(y-k)^2}{b}=1$

I have attached my work as a picture since it would take a long time to type in (I did it on Microsoft Office). Could anyone tell me if I am correct, and if not, where did I go wrong? Thanks!
• Mar 22nd 2010, 03:20 PM
masters
Quote:

Originally Posted by sleigh
I have to put the ellipse with the equation $\displaystyle ax^2+by^2+cx+dy+e=0$ into standard form, $\displaystyle \frac{(x-h)^2}{a}+\frac{(y-k)^2}{b}=1$

I have attached my work as a picture since it would take a long time to type in (I did it on Microsoft Office). Could anyone tell me if I am correct, and if not, where did I go wrong? Thanks!

Hi sleigh,

After completing the square in the 4th row, you factored incorrectly in the 5th row.

It should be:

$\displaystyle a\left(x+\frac{c}{2a}\right)^2+b\left(y+\frac{d}{2 b}\right)^2=-e+\frac{c^2}{4a}+\frac{d^2}{4b}$

Too bad the error came so early. Why on Earth would anyone want to do this, anyway?
• Mar 22nd 2010, 03:27 PM
sleigh
Quote:

Originally Posted by masters
Hi sleigh,

After completing the square in the 4th row, you factored incorrectly in the 5th row.

It should be:

$\displaystyle a\left(x+\frac{c}{2a}\right)^2+b\left(y+\frac{d}{2 b}\right)^2=-e+\frac{c^2}{4a}+\frac{d^2}{4b}$

Too bad the error came so early. Why on Earth would anyone want to do this, anyway?

Does this error make the entire thing wrong? Even though I factored wrong, I don't think this affects the rest of the problem because I added the correct amount to the other side, and I do not expand the binomials after I factored them.

I am doing this to program my calculator to put it into standard form and give all of the information. I have a program to do it in standard form, but I wanted it to work for general form, too. This way, for example, the x coorodinate of the vertex is $\displaystyle \frac{-c}{2a}$. This would work for the y coorodinate, $\displaystyle a^2$ term and $\displaystyle b^2$. I already did this for the parabola and circle, but I was unsure of this one. Thanks for your help!