Question

Find the center, vertices, and foci of the ellipse with equation 2x^2 + 6y^2 = 12

Answer 1

divide by 1 and start with
\[\frac{x^2}{6}+\frac{y^2}{2}=1\]

Answer 2

im kind of lost on how to do this can you explain on how to do to do this ??

Answer 3

general form is
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 4

center is \((h,k)\) so in your case it is \((0,0)\)

Answer 5

since \(6>2\) you know it looks like this

Answer 6

yes

Answer 7

how do i find the vertices

Answer 8

you have \(a^2=6\) and so \(a=\sqrt{6}\)

Answer 9

go \(\sqrt{6}\) units to the left and right of the center, which is \((0,0)\)

Answer 10

so it will be

Answer 11

no, right and left of \((0,0)\)

Answer 12

left is \((-\sqrt{6},0)\) and right is \((\sqrt{6},0)\)

Answer 13

ok im starting to understand :) ty

Answer 14

still need the foci right?

Answer 15

oh my picture was wrong, those were the vertices
should have looked like this