Prove that the six normals of an ellipsoid from an external point lie on a cubic curve.
Let $$x^2/a^2+y^2/b^2+z^2/c^2$$ be an ellipse .
Let $(\alpha,\beta,\gamma)$ be a point external to it from which six normals are drawn.
Clearly the feet of the normals are of the form $$ (a^2\alpha/a^2+k)etc$$
Let a plane $$ux+vy+wz+d=0$$ cut the ellipsoid at one of the feet of the normal, so we can prove that at least one nomral lie on a curve which is a cubic in K.
But how to show that this plane would intersect the ellipsoid at 6 places and those 6 points would be a cubic curve ?
Any other method would be appreciated.
Thank you.