Enveloping cone of an ellipsoid with vertex $P$ has parabolic sections by plane $z=0$. Locus of $P$?

Question

The section of the enveloping cone of the ellipsoid whose vertex is $P$, by the plane $z=0$ is a parabola. Find the locus of $P$.

The given ellipsoid is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ 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, $$ \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right) \left( \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}+\frac{z_1^2}{c^2}-1 \right)= \left( \frac{xx_1}{a^2}+\frac{yy_1}{b^2}+\frac{zz_1}{c^2}-1 \right)^2$$

This meets the plane $z=0$ then

$$ \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}-1 \right) \left( \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}+\frac{z_1^2}{c^2}-1 \right)= \left( \frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1 \right)^2$$

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

Answer 1

Rewrite the conic as

$$ \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{a^2} \left( \frac{y_1^2}{b^2}+\frac{z_1^2}{c^2}-1 \right) & -\frac{x_1 y_1}{a^2 b^2} & \frac{x_1}{a^2} \\ -\frac{x_1 y_1}{a^2 b^2} & \frac{1}{b^2} \left( \frac{x_1^2}{a^2}+\frac{z_1^2}{c^2}-1 \right) & \frac{y_1}{b^2} \\ \frac{x_1}{a^2} & \frac{y_1}{b^2} & -\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-\frac{z_1^2}{c^2} \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}=0$$

For parabola,

$$\det \begin{pmatrix} \frac{1}{a^2} \left( \frac{y_1^2}{b^2}+\frac{z_1^2}{c^2}-1 \right) & -\frac{x_1 y_1}{a^2 b^2} \\ -\frac{x_1 y_1}{a^2 b^2} & \frac{1}{b^2} \left( \frac{x_1^2}{a^2}+\frac{z_1^2}{c^2}-1 \right) \end{pmatrix} =0$$

The equation for $P$ is

$$ \frac{1}{a^2b^2} \left( \frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right) \left( \frac{x^2}{a^2}+\frac{z^2}{c^2}-1 \right)- \left( \frac{xy}{a^2 b^2} \right)^2=0$$

$$\frac{(z^2-c^2)}{a^2 b^2 c^2} \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right) =0$$

$$\fbox{$z^2=c^2$}$$

providing $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}>0$.

Answer 2

Here is a geometric derivation of the fact that the locus is the two planes with equations

$$z=c \ \ \text{and} \ \ z=-c\tag{1}$$

confirming the result found by @Ng Chung Tak.

Let it be clear (this is implicit in the solution attempt of the OP) that we consider an ellipsoid whose axes are the coordinate axes in a Euclidean setting.

Indeed, a necessary condition for the fact that the intersection of a cone with a plane is a parabola is the existence of a generatrix (aka "generating line") of the cone parallel to the section plane.

Let $(x_1,y_1,z_1)$ be the coordinates of the apex of the cone.

The problem being symmetrical with respect to plane $z=0$, we can assume $z_1>0$.

• if $z_1=c$, the horizontal line with equations $x_1y=y_1x, z=z_1$ is "responsible" for a parabolic section.

In the other cases, it is impossible:

• if $z_1>c$, no horizontal generatrix exists: it is evident by projecting the cone onto the vertical plane passing through the origin and the apex of the cone.

• if $0 \le z_1<c$ there are two horizontal generatrices, but projecting them onto the horizontal plane show that they are associated with vertical parabolic sections, not horizontal ones.

Remark 1: Using linear transformation $x=aY,y=bY,z=cZ$, one can convert this issue into the configuration of a cone tangent to the unit sphere $X^2+Y^2+Z^2=1$ which can be simpler conceptually (and technically if one has to do computations).

Remark 2: of course points $(0,0,\pm c)$ aren't part of the locus.