Vtyjkyuk

The given ellipsoid is $dfracx^2a^2+dfracy^2b^2+dfracz^2c^2=1$
And the vertex be $P(x_1,y_1,z_1)$
Therefore equation of enveloping cone of $P$ to this ellipsoid is $SS_1=T^2$.
That is,
$$
left(
fracx^2a^2+fracy^2b^2+fracz^2c^2-1
right)
left(
fracx_1^2a^2+fracy_1^2b^2+fracz_1^2c^2-1
right)=
left( fracxx_1a^2+fracyy_1b^2+fraczz_1c^2-1
right)^2$$

This meets the plane $z=0$ then

$$
left(
fracx^2a^2+fracy^2b^2-1
right)
left(
fracx_1^2a^2+fracy_1^2b^2+fracz_1^2c^2-1
right)=
left(
fracxx_1a^2+fracyy_1b^2-1
right)^2$$

I don't know after this some one plz help.

Rewrite the conic as

$$
beginpmatrix
x & y & 1
endpmatrix
beginpmatrix
frac1a^2
left( fracy_1^2b^2+fracz_1^2c^2-1 right) &
-fracx_1 y_1a^2 b^2 &
fracx_1a^2 \
-fracx_1 y_1a^2 b^2 &
frac1b^2
left( fracx_1^2a^2+fracz_1^2c^2-1 right) &
fracy_1b^2 \
fracx_1a^2 &
fracy_1b^2 &
-fracx_1^2a^2-fracy_1^2b^2-fracz_1^2c^2
endpmatrix
beginpmatrix
x \ y \ 1
endpmatrix=0$$

For parabola,

$$det
beginpmatrix
frac1a^2
left( fracy_1^2b^2+fracz_1^2c^2-1 right) &
-fracx_1 y_1a^2 b^2 \
-fracx_1 y_1a^2 b^2 &
frac1b^2
left( fracx_1^2a^2+fracz_1^2c^2-1 right)
endpmatrix
=0$$

The equation for $P$ is

$$
frac1a^2b^2
left( fracy^2b^2+fracz^2c^2-1 right)
left( fracx^2a^2+fracz^2c^2-1 right)-
left( fracxya^2 b^2 right)^2=0$$

$$frac(z^2-c^2)a^2 b^2 c^2
left( fracx^2a^2+fracy^2b^2+fracz^2c^2-1 right)
=0$$

$$fbox$z^2=c^2$$$

providing $dfracx^2a^2+dfracy^2b^2>0$.