A conicoid given by equation ax2+by2+cz2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0
is central if the system
⎩⎨⎧ax+hy+gz+u=0hx+by+fz+v=0gx+fy+cz+w=0
has a unique solution (x0,y0,z0). In this case the point (x0,y0,z0) is a center of the conicoid.
In our case, for a conicoid x2+2y2+2yz+2x+4y+8z+1=0 the system is the following:
⎩⎨⎧x+1=02y+z+2=0y+4=0
It is equivalent to
⎩⎨⎧x=−1y=−4z=6
Therefore, the conicoid is central with center at O′(−1,−4,6).
Let us find the new equation of the conicoid if the origin is shifted to its centre.
⎩⎨⎧x=x′−1y=y′−4z=z′+6
(x′−1)2+2(y′−4)2+2(y′−4)(z′+6)+2(x′−1)+4(y′−4)+8(z′+6)+1=0
(x′)2−2x′+1+2(y′)2−16y′+32+2y′z′+12y′−8z′−48+2x′−2+4y′−16+8z′+48+1=0
Therefore, the new equation of the conicoid if the origin is shifted to its centre is the following:
(x′)2+2(y′)2+2y′z′+16=0
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