Question

Show that the conicoid 2𝑥² + 2𝑦² + 𝑥𝑦 − 𝑦𝑧 + 𝑧𝑥 + 2𝑥 − 𝑦 + 5𝑧 + 1 = 0 is

central. Hence find its centre.

Answer 1

1] A conicoid, given by the equation

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

has the point $P(x_0,\ y_0,\ z_0)$ as a center if and only if;

$ax_0+hy_0+gz_0+u=0\\ hx_0+by_0+fz_0+v=0\\ gx_0+fy_0+cz_0+w=0$

Now, by comparing

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

with the given equation

$2𝑥^2 + 2𝑦^2 + 𝑥𝑦 − 𝑦𝑧 + 𝑧𝑥 + 2𝑥 − 𝑦 + 5𝑧 + 1 = 0$

we have;

$\displaystyle a=2,\ b=2,\ c=0,\ f=-\frac{1}{2},\ g=\frac{1}{2},\ h=\frac{1}{2},\ u=1,\ v=-\frac{1}{2},\ w=\frac{5}{2},\ d=1$

Thus,

$ax_0+hy_0+gz_0+u=0\\ hx_0+by_0+fz_0+v=0\\ gx_0+fy_0+cz_0+w=0$

$\Rightarrow$

$\displaystyle 4x_0+y_0+z_0=-2\\ x_0+4y_0-z_0=1\\ x_0-y_0+0z_0=-5$

$\displaystyle \Rightarrow x_0=-\frac{13}{5},\ y_0=\frac{12}{5},\ z_0=6$on solving the system of equations.

Thus, the center of the conicoid is $\displaystyle P\left(-\frac{13}{5},\ \frac{12}{5},\ 6\right)$.

2] A conicoid is called a central conicoid if it has a unique center.

To check if the given conicoid has a unique center, we need to show that the system of linear equations;

$\displaystyle 4x_0+y_0+z_0=-2\\ x_0+4y_0-z_0=1\\ x_0-y_0+0z_0=-5\\$

$\Rightarrow \begin{pmatrix} 4 & 1 & 1\\ 1 & 4 & -1\\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}= \begin{pmatrix} -2\\ 1\\ -5 \end{pmatrix}$ , in matrix form has a unique solution (exactly one solution) by showing that the determinant of the coefficient matrix is not equal to zero. That is we need to show that

$\begin{vmatrix} 4 & 1 & 1\\ 1 & 4 & -1\\ 1 & -1 & 0 \end{vmatrix}\neq 0$

Now,

$\begin{vmatrix} 4 & 1 & 1\\ 1 & 4 & -1\\ 1 & -1 & 0 \end{vmatrix}=-10\neq0$

Thus the coefficient matrix is invertible and $x_0,\ y_0,\ z_0$ are unique from the properties of matrix, with

$\displaystyle (x_0,\ y_0,\ z_0)=\left(-\frac{13}{5},\ \frac{12}{5},\ 6\right)$

showing that the conicoid center is unique.