Consider the tangent planes to the surface $S: \frac{x^2}{2}+y^2+z^2=1$ that passing through the point $P(1,1,1)$, then draw the perpendicular to the tangent plane from the centre of the surface $S$. What is the Locus of the foot of the perpendicular? $\\$ Here is my attempt: the tangent plane of the surface $S$ at one point $(x_{0},y_{0},z_{0})$ (the point $(x_{0},y_{0},z_{0})$ is on the surface) can be written as $\frac{x_{0}x}{2}+y_{0}y+z_{0}z=1$. For the point $(1,1,1)$ is on the plane, we obtain $\frac{x_{0}}{2}+y_{0}+z_{0}=1$. Note that one of normal vector of the plane $\frac{x_{0}x}{2}+y_{0}y+z_{0}z=1$ can be $(\frac{x_{0}}{2},y_{0},z_{0})$. So the equation of the perpendicular to the tangent plane from the centre of the surface $S$ can be written as $\begin{align} x = \frac{x_{0}t}{2} \\ y =y_0t \\ z =z_0t \end{align} $. Finally I obtain an equation of the Locus: $x^2=2xy+2yz+2xz$. Obviously, the Locus is a curve. How can I continue?
Find the Locus of the foot of the perpendicular.
analytic geometry
Best Answer
The (non planar) curve $(C)$ we are looking for is represented by blue dots on the image below. It is the intersection of 2 cones:
Fig. 1: The ellipsoid and the two cones. The curve is situated outside the ellipsoid but touches it in two points $(0,1,0)$ and $(0,0,1)$.
$$\frac12 x^2=xy+yz+xz\tag{1}$$
$$(\frac{x^2}{2}+y^2+z^2-1)\dfrac32 - (\frac{x}{2}+y+z-1)^2=0\tag{2}$$
using a classical formula (see here).
Indeed, considering the latter cone is equivalent to work with tangent planes.
I obtained $(C)$ "pointwise", by indirect means.
If I find a form of parametrized equation for $(C)$ (which should be with degree 4), I will write it down.
Remarks:
$$x^2+y^2+z^2-2(xy+xz+2yz)+2(x+2y+2z)-5=0$$
About the intersection of 2 cones, see here and as well here for the different cases with nice figures.
A "phantom" view on the same scene:)
Fig. 2.