Answer to Question #148541 in Analytic Geometry for Dhruv rawat

Question

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.

Answer 1

A conicoid given by equation $ax^2+by^2+cz^2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0$

is central if the system

$\begin{cases} ax+hy+gz+u=0 \\ hx+by+fz+v=0 \\ gx+fy+cz+w=0 \end{cases}$

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

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

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

It is equivalent to

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

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.

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

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

$(x')^2-2x'+1+2(y')^2-16y'+32+2y'z'+12y'-8z'-48+2x'-2+4y'-16+8z'+48+1=0$

Therefore, the new equation of the conicoid if the origin is shifted to its centre is the following:

$(x')^2+2(y')^2+2y'z'+16=0$