Answer to Question #293622 in Analytic Geometry for Adam

Question

Examine which of the following conicoids are central and which are non-central. Also determine which of the central conicoids have centre at the origin.

(i) 𝑥² + 𝑦² + 𝑧² + 4𝑥 + 3𝑦 − 𝑧 = 0

(ii) 2𝑥² − 𝑦² − 𝑧² + 𝑥𝑦 + 𝑦𝑧 − 𝑧𝑥 = 1

(iii) 𝑥² + 𝑦² − 𝑧² − 2𝑥𝑦 − 3𝑦𝑧 − 6𝑧𝑥 + 𝑥 − 2𝑦 + 5𝑧 + 4 = 0

Answer 1

A conicoid given by equation

$$ax^2 +by^2 +cz^2 +2fyz+2gxz+2hxy$$

$$+2ux+2vy+2wz+d=0$$

has a point $(x_0 ,y_0 ,z_0 )$ as a center if

$$\begin{cases} ax_0+hy_0+gz_0+u=0 \\ hx_0+by_0+fz_0+v=0 \\ gx_0+fy_0+cz_0+w=0 \end{cases}$$

(i)

$a=1, b=1, c=1,$

$f=g=h=0,$

$u=2,v=3/2, w=-1/2$

Then

$$\begin{cases} x_0+2=0 \\ y_0+3/2=0 \\ z_0-1/2=0 \end{cases}$$

The central conicoid has a center $(-2,-3/2,1/2).$

(ii)

$a=2, b=-1, c=-1,$

$f=1/2,g=-1/2,h=1/2,$

$u=v=w=0$

Then

$$\begin{cases} 2x_0+\dfrac{1}{2}y_0-\dfrac{1}{2}z_0=0 \\ \\ \dfrac{1}{2}x_0-y_0+\dfrac{1}{2}z_0=0 \\ \\ -\dfrac{1}{2}x_0+\dfrac{1}{2}y_0-z_0=0 \end{cases}$$

The central conicoid has a center $(0,0,0).$

(iii)

$a=1, b=1, c=-1,$

$f=-3/2,g=-3,h=-1,$

$u=1/2,v=-1,w=5/2$

Then

$$\begin{cases} x_0-y_0-3z_0+\dfrac{1}{2}=0 \\ \\ -x_0+y_0-\dfrac{3}{2}z_0-1=0 \\ \\ -3x_0-\dfrac{3}{2}y_0-z_0+\dfrac{5}{2}=0 \end{cases}$$

The central conicoid has a center $(\dfrac{49}{162},\dfrac{92}{81},-\dfrac{1}{9}).$