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. 

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 $(2,2,-1)$.
