Conic Sections

Sw0rDz · #1 $Old$ November 4th, 2009, 04:38 PM

Problem: $4x^2 + 4x +y^2 = 0$

I graph it and I get an elipse, but I cant figure out how to get it into the form:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$

I've using the form:

$b^2x^x + a^2y^2 = a^2b^2$

and setting:

$a^2b^2 = -4x$

But, nothing came good out of that.

Any suggestions?

skeeter · #2 $Old$ November 4th, 2009, 05:01 PM

Quote:

Originally Posted by Sw0rDz $View Post$

Problem: $4x^2 + 4x +y^2 = 0$

I graph it and I get an elipse, but I cant figure out how to get it into the form:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$

I've using the form:

$b^2x^x + a^2y^2 = a^2b^2$

and setting:

$a^2b^2 = -4x$

But, nothing came good out of that.

Any suggestions?

$4x^2 + 4x +y^2 = 0$

$4\left(x^2 + x + \frac{1}{4}\right) + y^2 = 1$

$4\left(x + \frac{1}{2}\right)^2 + y^2 = 1$

$\frac{\left(x + \frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2} + \frac{y^2}{1^2} = 1$

Sw0rDz · #3 $Old$ November 4th, 2009, 05:15 PM

Quote:

Originally Posted by skeeter $View Post$

$4x^2 + 4x +y^2 = 0$

$4\left(x^2 + x + \frac{1}{4}\right) + y^2 = 1$

$4\left(x + \frac{1}{2}\right)^2 + y^2 = 1$

$\frac{\left(x + \frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2} + \frac{y^2}{1^2} = 1$

Awh!! Thankyou!