Question #148541
Show that the conicoid x^2+2y^2+2yz+2x+4y+8z+1=0 is central. Find the new equation of the conicoid if the origin is shifted to its centre.
1
Expert's answer
2020-12-07T10:53:27-0500

A conicoid given by equation ax2+by2+cz2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0ax^2+by^2+cz^2+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\begin{cases} ax+hy+gz+u=0 \\ hx+by+fz+v=0 \\ gx+fy+cz+w=0 \end{cases}


has a unique solution (x0,y0,z0)(x_0,y_0,z_0). In this case the point (x0,y0,z0)(x_0,y_0,z_0) is a center of the conicoid.


In our case, for a conicoid x2+2y2+2yz+2x+4y+8z+1=0x^2+2y^2+2yz+2x+4y+8z+1=0 the system is the following:


{x+1=02y+z+2=0y+4=0\begin{cases} x+1=0 \\ 2y+z+2=0 \\ y+4=0 \end{cases}


It is equivalent to


{x=1y=4z=6\begin{cases} x=-1 \\ y=-4 \\ z=6 \end{cases}


Therefore, the conicoid is central with center at O(1,4,6)O'(-1,-4,6).


Let us find the new equation of the conicoid if the origin is shifted to its centre.


{x=x1y=y4z=z+6\begin{cases} x=x'-1 \\ y=y'-4 \\ z=z'+6 \end{cases}


(x1)2+2(y4)2+2(y4)(z+6)+2(x1)+4(y4)+8(z+6)+1=0(x'-1)^2+2(y'-4)^2+2(y'-4)(z'+6)+2(x'-1)+4(y'-4)+8(z'+6)+1=0


(x)22x+1+2(y)216y+32+2yz+12y8z48+2x2+4y16+8z+48+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+2yz+16=0(x')^2+2(y')^2+2y'z'+16=0




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