A conicoid given by equation
a x 2 + b y 2 + c z 2 + 2 f y z + 2 g x z + 2 h x y ax^2 +by^2 +cz^2 +2fyz+2gxz+2hxy a x 2 + b y 2 + c z 2 + 2 f yz + 2 gx z + 2 h x y
+ 2 u x + 2 v y + 2 w z + d = 0 +2ux+2vy+2wz+d=0 + 2 ux + 2 v y + 2 w z + d = 0 has a point ( x 0 , y 0 , z 0 ) (x_0 ,y_0 ,z_0 ) ( x 0 , y 0 , z 0 )
{ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 g x 0 + f y 0 + c z 0 + w = 0 \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} ⎩ ⎨ ⎧ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 g x 0 + f y 0 + c z 0 + w = 0
(i)
a = 1 , b = 1 , c = 1 , a=1, b=1, c=1, a = 1 , b = 1 , c = 1 ,
f = g = h = 0 , f=g=h=0, f = g = h = 0 ,
u = 2 , v = 3 / 2 , w = − 1 / 2 u=2,v=3/2, w=-1/2 u = 2 , v = 3/2 , w = − 1/2
Then
{ x 0 + 2 = 0 y 0 + 3 / 2 = 0 z 0 − 1 / 2 = 0 \begin{cases}
x_0+2=0 \\
y_0+3/2=0 \\
z_0-1/2=0
\end{cases} ⎩ ⎨ ⎧ x 0 + 2 = 0 y 0 + 3/2 = 0 z 0 − 1/2 = 0 The central conicoid has a center ( − 2 , − 3 / 2 , 1 / 2 ) . (-2,-3/2,1/2). ( − 2 , − 3/2 , 1/2 ) .
(ii)
a = 2 , b = − 1 , c = − 1 , a=2, b=-1, c=-1, a = 2 , b = − 1 , c = − 1 ,
f = 1 / 2 , g = − 1 / 2 , h = 1 / 2 , f=1/2,g=-1/2,h=1/2, f = 1/2 , g = − 1/2 , h = 1/2 ,
u = v = w = 0 u=v=w=0 u = v = w = 0
Then
{ 2 x 0 + 1 2 y 0 − 1 2 z 0 = 0 1 2 x 0 − y 0 + 1 2 z 0 = 0 − 1 2 x 0 + 1 2 y 0 − z 0 = 0 \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} ⎩ ⎨ ⎧ 2 x 0 + 2 1 y 0 − 2 1 z 0 = 0 2 1 x 0 − y 0 + 2 1 z 0 = 0 − 2 1 x 0 + 2 1 y 0 − z 0 = 0 The central conicoid has a center ( 0 , 0 , 0 ) . (0,0,0). ( 0 , 0 , 0 ) .
(iii)
a = 1 , b = 1 , c = − 1 , a=1, b=1, c=-1, a = 1 , b = 1 , c = − 1 ,
f = − 3 / 2 , g = − 3 , h = − 1 , f=-3/2,g=-3,h=-1, f = − 3/2 , g = − 3 , h = − 1 ,
u = 1 / 2 , v = − 1 , w = 5 / 2 u=1/2,v=-1,w=5/2 u = 1/2 , v = − 1 , w = 5/2
Then
{ x 0 − y 0 − 3 z 0 + 1 2 = 0 − x 0 + y 0 − 3 2 z 0 − 1 = 0 − 3 x 0 − 3 2 y 0 − z 0 + 5 2 = 0 \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} ⎩ ⎨ ⎧ x 0 − y 0 − 3 z 0 + 2 1 = 0 − x 0 + y 0 − 2 3 z 0 − 1 = 0 − 3 x 0 − 2 3 y 0 − z 0 + 2 5 = 0 The central conicoid has a center ( 49 162 , 92 81 , − 1 9 ) . (\dfrac{49}{162},\dfrac{92}{81},-\dfrac{1}{9}). ( 162 49 , 81 92 , − 9 1 ) .
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