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