# Conic section

• Sep 3rd 2008, 07:38 AM
OnMyWayToBeAMathProffesor
Conic section
I know this is a conic section and that it is a circle, I can get that much but I have forgotten how to simplify it down to a variation of the standard formula, $\displaystyle x^2+y^2=r^2$. I would love for someone to refresh my memory, Thank you.
$\displaystyle 4x^2+y^2-8x+4y+4=0$
• Sep 3rd 2008, 08:02 AM
Chris L T521
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
I know this is a conic section and that it is a circle, I can get that much but I have forgotten how to simplify it down to a variation of the standard formula, $\displaystyle x^2+y^2=r^2$. I would love for someone to refresh my memory, Thank you.
$\displaystyle 4x^2+y^2-8x+4y+4=0$

$\displaystyle 4x^2+y^2-8x+4y+4=0$

Group the x and y terms together:

$\displaystyle (4x^2-8x)+(y^2+4y+4)=0\implies 4(x^2-2x)+(y^2+4y+4)=0$

Complete the square for the x term. The equation then becomes:

$\displaystyle 4(x^2-2x+1)+(y^2+4y+4)=4$

Now factor:

$\displaystyle 4(x-1)^2+(y+2)^2=4$

The equation now becomes $\displaystyle \color{red}\boxed{(x-1)^2+\frac{(y+2)^2}{4}=1}$

This is an ellipse with the major axis along the y-axis.

I hope this makes sense! (Sun)

--Chris
• Sep 3rd 2008, 08:18 AM
OnMyWayToBeAMathProffesor
THANK YOU very much, I knew it was something simple like that. It has been so long since i did a problem like that.