0

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.

Rodrigo de Azevedo
  • 1
bharathi b
  • 1
  • Recall the relationship that the cone and plane must have for the section to be a parabola. – amd Aug 12 '18 at 08:23

2 Answers2

2

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

Ng Chung Tak
  • 18,990
  • I just posted a geometrical solution. A very tiny detail in the last sentence: "providing" $\to$ "provided". – Jean Marie Sep 03 '20 at 05:09
  • @Ng Chung Tak How did you conclude that for a parabola that determinant should be zero? – Arthur Feb 02 '23 at 09:17
  • 1
    @Franklin Please refer to Matrix notation and Discriminant under General Cartesian form of Conic Section here. – Ng Chung Tak Feb 02 '23 at 09:24
  • @NgChungTak That was truly a great enlightenment I had. Thank you! Just one more thing : how did you obtain from $\frac{(z^2-c^2)}{a^2 b^2 c^2} \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right) =0$ only $z^2=c^2$ is possible. Is it because $x^2/a^2+y^2/b^2+z^2/c^2=1$, isn't possible ? If so,why ? – Arthur Feb 02 '23 at 09:54
  • That's usual trick of switching dummy variables in coordinate geometry. $(x_1,y_1,z_1)$ satisfies $\det \ldots =0$ which is just the equation of $P$. Finally, the radiant point of the cone should be outside the ellipsoid. – Ng Chung Tak Feb 02 '23 at 12:03
  • @NgChungTak Thank you! So the real point is that since $(x_1,y_1,z_1)$ is the vertex of the considered cone and hence can't lie on the ellipsoid, right? – Arthur Feb 02 '23 at 16:09
0

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.

Jean Marie
  • 81,803