Plane through feet of normals to ellipsoid

Question

The problem

If $P,Q,R,P',Q',R'$ be the feet of six normals drawn from a point to the ellipsoid

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1=0$$ and the plane $PQR$ is represented by $lx+my+nz-p=0$, show that the plane $P'Q'R'$ is represented by $$\frac{x}{a^2l} + \frac{y}{b^2m} + \frac{z}{c^2n} + \frac{1}{p} = 0$$

What I have done so far:

If the original point is $(\alpha, \beta, \gamma)$, then any foot of the perpendicular is of the form: $$(\frac{\alpha a^2}{\lambda+a^2}, \frac{\beta b^2}{\lambda+b^2}, \frac{\gamma c^2}{\lambda+c^2})$$ where $\lambda$ is parameter. Putting this in the ellipsoid, I get a six degree equation as: $$\frac{\alpha^2 a^2}{(\lambda+a^2)^2}+ \frac{\beta^2 b^2}{(\lambda+b^2)^2}+ \frac{\gamma^2 c^2}{(\lambda+c^2)^2} - 1 = 0$$

Assuming the other plane to be $l'x + m'y + n'z +p'=0$, I put the values of the parametric point and multiply the two planes. $$(lx+my+nz-p)(l'x+m'y+n'y-p')=0$$ or $$(l\frac{\alpha a^2}{\lambda+a^2}+m\frac{\beta b^2}{\lambda+b^2}+n\frac{\gamma c^2}{\lambda+c^2}-p) (l'\frac{\alpha a^2}{\lambda+a^2}+m'\frac{\beta b^2}{\lambda+b^2}+n'\frac{\gamma c^2}{\lambda+c^2}-p')=0$$

Comparing coefficients, this should be equal to the parametric ellipsoid equation. But to arrive at the result, I have to make the $\alpha^2\beta^2$ terms zero. How do I do that? Is my approach correct?

Answer 1

There is a related question on this site -- 6 Normals of an Ellipsoid -- whose comments point to three textbooks containing worked answers of the problem.

The answers use the same technique you've outlined, including the claim that the product of the plane equations should yield the equation of the ellipsoid.

That's not correct .. the product of two plane equations will yield the equation of a degenerate quadric consisting of two planes. So I don't think the approach is correct.

I mocked up a 2d version of this in Geogebra to see what was going on. Remarkably, the equations matched (up to a multiplicative constant) in their $x^2$ and $y^2$ coefficients, so the approach ends up getting the desired answer even though it's either wrong or requires further justification.

Update: Some observations and references that may be useful to understanding or solving the problem:

Aliyev, Apollonius Problem And Caustics Of An Ellipsoid gives a useful discussion of the 2D and 3D problem. The 2D version was posed and solved by a dude named Apollonius back in B.C. times.

As shown in the above figure (from the paper) the feet of the four normals from $A$ are the intersection of the ellipse with a rectangular hyperbola running through $A$ and the center of the ellipse.

In the theory of conics, the entire set of conics running through the four points is called a pencil. Letting $e,h$ be the ellipse and the hyperbola, the equation for any conic in the pencil is a linear combination of the equations for $e$ and $h$. Any pair of two lines, each running through a different pair of feet, can be multiplied to represent a degenerate conic that is a member of the pencil, therefore a weighted sum of $e$ and $h$. But the equation for $e$ uses terms in $x^2,y^2,1$ and the equation for $h$ uses terms in $xy, x, y$. The fact that the terms are disjoint takes care of making the $\alpha^2\beta^2$ terms in the OP collectively zero.

Can this work in 3D? In that case there is a cone that contains the normals (using terms in $xy,yz,zx,x,y,z$), a cone that goes through the origin and contains the feet (using terms in $xy,yz,zx$), and these may somehow take the place of $h$. I've observed in Geogebra that the product of any pair of planes generated by six feet will be a linear combination of the ellipsoid and a quadric with terms in $xy,yz,zx,x,y,z$ that runs through $A$, the origin, and the six feet. But it doesn't look as straightforward as the 2D case.

For more info on the aforementioned cones see Henderson, The Cone Of The Normals And An Allied Cone For Central Surfaces Of The Second Degree.

Perhaps the most promising tip I can give is in Baker, Principles of Geometry, Vol 3, pgs 104-105. Exs. 1-4 culminate in a proof of the form of the equations for the two planes in the OP. Good luck!