Posted by **pedro** on Tuesday, July 10, 2012 at 11:45am.

Find the vertices and foci of this ellipse 4x^2+16y^2+64x+64y+256=0

- algebra -
**Steve**, Tuesday, July 10, 2012 at 1:09pm
4x^2+16y^2+64x+64y+256=0

4x^2+64x+256 + 16y^2+64y+64 = -256

4(x+8)^2 + 16(y+2)^2 = -256 + 256 + 64

(x+8)^2 + 4(y+2)^2 = 16

(x+8)^2/16 + (y+2)^2/4 = 1

h=-8 k=-2

a=4 b=2 c=√12

center: (-8,-2)

vertices: (-12,-2) (-4,-2)

foci: (-8-√12,-2) (-8+√12,-2)

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