draw the ellipse and find the center, eccentricity, and foci of x^2+4y^2-6x-24y+41=0

Question

anonymous · Answer

here is a nice picture mertsj will give an explanation, which i imagine involves completing the square to write in standard form http://www.wolframalpha.com/input/?i=ellipse++x^2%2B4y^2-6x-24y%2B41%3D0

mertsj · Answer

$$x^2-6x+9+4(y^2+6y+9)=-41+45$$
$$\frac{(x-3)^2}{4}+\frac{(y+3)^2}{1}=1$$
Center (3,-3) 
a=2, b=1, c=sqrt3 e=sqrt3/2

anonymous · Answer

useful tool that "completing the square"

mertsj · Answer

yep

mertsj · Answer

But I see I got a sign wrong..  It's negative 24 y

mertsj · Answer

So center is actually (3,3)

mertsj · Answer

Wish there was some way to correct a post.

anonymous · Answer

how about the eccentricity?

anonymous · Answer

the foci,rather,.,.not the eccentricity,.,.

mertsj · Answer

Foci are positive and negative c units to the right and left of the center.

c=sqrt (a^2-b^2)