Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\right)\\
\left(-\frac{6\sqrt{13}}{13}, -\frac{6\sqrt{13}}{13}, \frac{3\sqrt{13}}{13}\right)$$
I know the normal is $(x, 4y, z)$. What step comes after this?
Thanks in advance. 
 A: Obviously, the points that you mentioned are not the answer, since they don't even belong to the ellipsoid.
The points where the tangent plane is perpendicular to the given line are $\pm\left(\sqrt6,\sqrt{\frac38},-\sqrt{\frac32}\right)$, since these are the points of the ellipsoid such that the gradient of $x^2+4y^2+z^2$ is a multiple of $(2,2,-1)$.
A: Projective geometry to the rescue. 
The points on the line (in homogeneous coordinates) are 
$${\bf q} =\left[ \matrix{ {\bf r} + {\bf e}\, t \\ 1}\right]  =\left[ \matrix{ \pmatrix{2\\1\\3}+\pmatrix{2\\2\\-1} t \\ 1}\right] $$
The planes normal to the line (in homogeneous coordinates) are
$${\bf w} =\left[ \matrix{ {\bf e} \\ -({\bf r}+{\bf e}\,t)\cdot{\bf e}} \right] =\left[ \matrix{ \pmatrix{2\\2\\-1} \\ -3 (3 t+1) } \right] $$
The ellipsoid coefficients are
$$ {\rm C} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -9 \end{bmatrix}$$
A property of homogeneous coordinates is that all the equation of the ellipsoid is defined by the locus of points $${\bf p}^\top {\rm C}\,{\bf p} =0$$ where ${\bf p}=\pmatrix{x&y&z&1}^\top$ is the point coordinates. 
Another property is that if ${\bf p}$ is a point on the ellipse, then ${\bf w} = {\rm C}\, {\bf p}$ is the coordinates if the plane tangent to the ellipse (through the point). 
These two facts lead to the following lemma:
An ellipsoid is defined by the envelope of tangent planes by the following equation $${\bf w}^\top {\rm C}^{-1}\,{\bf w} =0$$
This leads to the equation 
$$\matrix{ -9 t^2 -6 t+5 =0 & t = \cases{ 
   -\frac{1}{3} + \frac{\sqrt{6}}{3} \\ 
   -\frac{1}{3} - \frac{\sqrt{6}}{3} } }$$
which yield two tangent planes 
$$ \matrix{
  {\bf w}_1 =\left[ \matrix{ \pmatrix{2\\2\\-1}\\-3\sqrt{6}} \right] &
  {\bf w}_2 =\left[ \matrix{ \pmatrix{2\\2\\-1}\\3\sqrt{6}} \right]
}$$
And the corresponding points ${\bf p}_i = {\rm C}^{-1} {\bf w}_i$
$$ \matrix{
  {\bf p}_1 =\left[ \matrix{ \pmatrix{2\\ \frac{1}{2} \\-1}\\ \frac{\sqrt{6}}{3} } \right] &
  {\bf p}_2 =\left[ \matrix{ \pmatrix{2\\ \frac{1}{2} \\-1}\\ -\frac{\sqrt{6}}{3} } \right]
}$$
In cartesian coordinates the above points are is
$$ \matrix{
{\bf r}_1 = \pmatrix{ \sqrt{6} \\ \frac{\sqrt{6}}{4} \\ -\frac{\sqrt{6}}{2} } & {\bf r}_2 = \pmatrix{ -\sqrt{6} \\ -\frac{\sqrt{6}}{4} \\ \frac{\sqrt{6}}{2} }
}$$
Confirmation
Using GeoGebra

Points d and e are the tangent points. The coordinates of $$e = \pmatrix{-2.4494897427831780 \\-0.61237243569579452 \\ 1.2247448713915890}$$ and $$d = \pmatrix{2.4494897427831780 \\ 0.61237243569579452 \\ -1.2247448713915890}$$
match the solution above.
A: The normal to the plane must be parallel to the given line, hence
$$(x,4y,z)=\lambda(2,2,-1).$$
Combining with the equation of the ellipsoid,
$$x^2+4y^2+z^2=9,$$
we get
$$2^2\lambda^2+4\frac1{2^2}\lambda^2+(-1)^2\lambda^2=6\lambda^2=9,$$ and solve for $\lambda$.
$$(x,y,z)=\pm\sqrt{\frac32}\left(2,\frac12,-1\right).$$
