Find all the points where the tangent plane is perpendicular to a given vector.

Question

Given an ellipsoid : x^2+4y^2+z^2=9 Find the tangent plane of the ellipsoid at (x_0,y_0,z_0) I found the equation for the tangent plane to be: 2x(x-x_0)+8y(y-y_0)+2z(z-z_0)=0 but I need to find every point where this plane is perpendicular to the vector <-4,8,-2>. I understand what is going on geometrically and that there should be two points ( multiplying the vector by -1) that satisfy these conditions but I'm not sure how to find the points. Can someone please point me in the right direction? (pun intended)

Answer 1

Being given a general quadric with implicit equation:

$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gx+2hy+2iz+k=0,$$

the key point is that the tangent plane to this quadric in $(x',y',z')$ is obtained by "polarization", i.e., by looking at the homogeneous version of (1). This homogeneous form is obtained by introducing a 4th variable $t$, atop variables $x,y,z$, in such a way that if a term is only 1st degree, it is multiplied by $t$, and if it is a constant, it is multiplied by $t^2$: thus, all terms become of degree 2 (whence the name "homogenization"), in this way:

$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gtx+2hty+2itz+kt^2=0$$

Remark: if $t=1$, one finds back (1).

Then the method amounts to build the bilinear form associated with this quadratic form in 4 variables:

$\tag{2}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(t'x+x't)+h(t'y+ty')+i(t'z+tz')+ktt'=0.$

Remark: if all the "primes" are suppressed, we are back to equation (2).

Then make $t=t'=1$ in (2),

$$\tag{3}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(x+x')+h(y+y')+i(z+z')+k=0,$$

We have obtained in (3) the equation of the tangent plane to the quadric at point $(x',y',t')$.

Remark: this techniques can be justified in the framework of projective geometry.

Applying this to our case, we get, for the tangent plane at $(x_0,y_0,z_0)$ is

$$xx_0+4yy_0+zz_0=9$$

This is the equation of a plane with normal vector

$$\pmatrix{x_0\\4y_0\\z_0} \text{which is desired to be proportional to} \pmatrix{-4\\8\\-2}.$$

Thus, we must look for a coefficient $\alpha$ such that:

$$\tag{4}\pmatrix{x_0\\4y_0\\z_0}=\pmatrix{-4\alpha\\8\alpha\\-2\alpha}$$

Point $(x_0,y_0,z_0)$ is a point of the ellipsoid if and only if:

$$(-4\alpha)^2+4(2\alpha)^2+(-2\alpha)^2=9$$

i.e., $$\alpha^2=\frac{9}{36}=\frac{1}{4} \ \ \Leftrightarrow \ \ \alpha=\pm \frac{1}{2}$$

Whence the two solution points by plugging these opposite values of $\alpha$ in (4):

$$\pmatrix{x_0\\y_0\\z_0}=\pm \frac{1}{2} \pmatrix{-4\\2\\-2}=\pm \pmatrix{-2\\1\\-1}.$$

Answer 2

Let $g = x^2+4y^2+z^2-9$ then $g^{-1}(0)=E$ where $E$ denoted the ellipsoid. Recall then that $\nabla g(p)$ is a normal vector for the plane where $p = (x_0,y_0,z_0)$. Thus if you want the plane to be perpendicular to $\vec{v} = \langle -4,8,-2\rangle$ then you just need all $p$ s.t $\nabla g(p)= \lambda \vec{v}$.

Answer 3

The gradient vector $\nabla F = [F_x, F_y, F_z]^T$ is normal (perpendicular) to the surface of the ellipsoid $F(x, y, z) = 0 $ and therefore will be perpendicular to the tangent plane.

The gradient for the given ellipsoid is given by

$ \nabla F = [2 x, 8 y, 2 z]^T $

We want this vector to be proportional to the given vector $V = [-4, 8, -2]^T $

That is, we want

$ \begin{bmatrix} 2 x \\ 8 y \\ 2 z \end{bmatrix} = \alpha \begin{bmatrix} -4 \\ 8 \\ -2 \end{bmatrix} $

It follows that

$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \alpha \begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix} $

Now substitute this point into the ellipsoid equation to determine $\alpha$.

This yields

$ \alpha^2 ( 4 + 1 + 1 ) = 9 $

So $\alpha = \pm \sqrt{ \dfrac{3}{2} }$

Hence, the required points are the following two points

$ (x_1, y_1, z_1 ) = \sqrt{ \dfrac{3}{2} } (-2, 1, -1) $ and $(x_2,y_2, z_2) = - \sqrt{ \dfrac{3}{2} } ( -2, 1, -1) $