Question

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

Answer 1

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

Answer 2

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

Answer 3

useful tool that "completing the square"

Answer 4

yep

Answer 5

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

Answer 6

So center is actually (3,3)

Answer 7

Wish there was some way to correct a post.

Answer 8

how about the eccentricity?

Answer 9

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

Answer 10

Foci are positive and negative c units to the right and left of the center.

c=sqrt (a^2-b^2)