Find the center of the ellipsoid

Question

x^2+4y+2z^2+2x-4y+z = 20

I'm having trouble factoring to get the center points. Can someone please help. This is where I'm stuck.

(x^2+2x+1)+4(y^2-y+1/4)+2(x^2+1/2z+1/16)=20+1+1+1/8

(x+1)^2+
Now I don't know how to get the y and z plane factored.

Answer 1

x^2+4??????y^2 ?????+2z^2+2x-4y+z = 20

if so
x^2 + 2 x = -(+4y^2+2z^2-4y+z) + 20

x^2 + 2 x + 1 = -(+4y^2+2z^2-4y+z) + 21

(x+1)^2 = -(+4y^2+2z^2-4y+z) + 21

4 y^2 - 4 y = -(x+1)^2 -2 z^2 -z + 21

y^2 - y =(1/4)[-(x+1)^2 -2 z^2 -z + 21]

y^2-y+1/4=(1/4)[-(x+1)^2 -2 z^2 -z + 22]

4(y-1/2)^2=[-(x+1)^2 -2 z^2 -z+ 22]

2 z^2 + z = -4(y-1/2)^2 -(x+1)^2 +22

z^2 + (1/2)z =-2 (y-.5)^2 -.5(x+1)^2 +22

z^2+.5z+1/16=-2 (y-.5)^2-.5(x+1)^2+22 1/16

(z +1/4)^2 +2(y-1/2)^2 + 1/2(x+1^2 = 22 1/16

check my arithmetic !!!

Answer 2

anyway that would put the center at
x = -1
y = 1/2
z = -1/4

Answer 3

Hmmm. assuming some typos, I get

(x^2+2x+1) + 4(y^2-y+1/4)+2(z^2+z/2+1/16) = 20+1+1+1/8

(x-1)^2 + 4(y-1/2)^2 + 2(z+1/4)^2 = 177/8

At this point we know that the center is at (1,1/2,1/4).

We can go ahead and divide by 177/8 to get it into standard form, but it won't affect the center.

(x-1)^2/(177/8) + (y-1/2)^2/(177/2) + (z-1/4)^2/(177/4) = 1

Answer 4

Oww! I messed up the signs. Damon's values are correct.