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So, $x^2+y^2+4z^2-2x-16z+12=0$ is an ellipsoid-shaped surface. I am asked to find the points on the surface that has a tangent plane which is parallel to the XZ-plane (or y = 0). I haven't really found anything that can help. Can you help giving me hints/answers and explaining it thoroughly so I can understand? Thank you!

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Hint. A normal vector to the tangent plane at $(x,y,z)$ is given by: $$\begin{pmatrix}2x-2\\2y\\8z-16\end{pmatrix}.$$ A normal vector to the $XZ$-plane is given by: $$\begin{pmatrix}0\\1\\0\end{pmatrix}$$

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  • $\begingroup$ I see we were thinking alike. Although you were nicer and gave away the gradient. $\endgroup$ – Faraad Armwood May 16 '17 at 19:35
  • $\begingroup$ Indeed, I just saw that! I believe this is more or less the canonical approach to this kind of problem. :) $\endgroup$ – C. Falcon May 16 '17 at 19:37
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    $\begingroup$ Oh definitely. I wouldn't do it any other way. In any case, there is still some work fro the OP to do. $\endgroup$ – Faraad Armwood May 16 '17 at 19:37
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$\textbf{Hint}$: You want to be parallel to the plane $y = 0$ and so the normal vector of the plane must be parallel to $\vec{v} = (0,1,0)$. How do you get a normal vector from an equation $f(x,y,z) = 0$? Use the gradient.

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The general approach to this type of problem is to compute the gradient of the surface and find the points at which it is parallel to the desired tangent plane normal, which in this case is $(0,1,0)^T$. So for this problem, you’d be looking for points on the surface at which the $x$- and $z$- components of the gradient both vanish.

For this problem, though, you can instead take advantage of the geometry. Since there are only squared and linear terms in the variables, the principal axes are parallel to the coordinate axes. The tangent planes at the ends of the principal axes are thus parallel to the coordinate planes. The center of the ellipsoid can be found by setting the partial derivatives of its equation to zero and solving, which yields the point $(1,0,2)$. The principal axis that’s perpendicular to the $x$-$z$ plane is then the line $x=1$, $z=2$ and the problem thus reduces to plugging these values of $x$ and $z$ into the equation of the surface and solving for $y$.

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