1. Tangent Planes Parallel to a Given Plane.
Find the points on the surface
\begin{equation*}
x^2 + y^2 + 3z^2 = 1
\end{equation*}
at which the tangent plane is parallel to the plane \(-x - 5y - 2z = -3\text{.}\)
Solution.
Multiplying by \(-1\text{,}\) the given plane is \(x + 5y + 2z = 3\text{,}\) with normal vector \(\mathbf{n} = \langle 1, 5, 2\rangle\text{.}\) The tangent plane to the ellipsoid is parallel to this plane exactly at the points of the ellipsoid that are closest to and farthest from it, so we minimize and maximize the distance to the plane subject to the constraint of lying on the ellipsoid.
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Choose a point on the plane, say \(P(3,0,0)\text{.}\) For a point \(S(x,y,z)\) on the ellipsoid, the distance from \(S\) to the plane is
\begin{equation*}
\frac{\overrightarrow{PS}\cdot\mathbf{n}}{\lvert\mathbf{n}\rvert}
= \frac{\langle x-3,\, y,\, z\rangle\cdot\langle 1,5,2\rangle}
{\lvert\mathbf{n}\rvert}.
\end{equation*}
Since \(\lvert\mathbf{n}\rvert\) is constant, it suffices to optimize the numerator
\begin{equation*}
f(x,y,z) = x - 3 + 5y + 2z
\end{equation*}
subject to the constraint
\begin{equation*}
g(x,y,z) = x^2 + y^2 + 3z^2 - 1 = 0.
\end{equation*}
The Lagrange condition \(\nabla f = \lambda\, \nabla g\) gives
\begin{equation*}
\langle 1, 5, 2\rangle = \lambda \langle 2x,\, 2y,\, 6z\rangle,
\end{equation*}
so
\begin{equation*}
x = \frac{1}{2\lambda}, \qquad
y = \frac{5}{2\lambda}, \qquad
z = \frac{1}{3\lambda}.
\end{equation*}
Substituting into the constraint \(x^2 + y^2 + 3z^2 = 1\text{,}\)
\begin{align*}
\frac{1}{4\lambda^2} + \frac{25}{4\lambda^2} + \frac{3}{9\lambda^2} \amp= 1\\
\frac{3 + 75 + 4}{12\lambda^2} \amp= 1\\
\frac{1}{\lambda^2} \amp= \frac{12}{82},
\end{align*}
so that
\begin{equation*}
\frac{1}{\lambda} = \pm 2\sqrt{\frac{3}{82}}.
\end{equation*}
Therefore
\begin{equation*}
x = \pm\sqrt{\frac{3}{82}}, \qquad
y = \pm 5\sqrt{\frac{3}{82}}, \qquad
z = \pm\sqrt{\frac{2}{123}},
\end{equation*}
and the two points are
\begin{equation*}
S_1\!\left(\sqrt{\tfrac{3}{82}},\; 5\sqrt{\tfrac{3}{82}},\; \sqrt{\tfrac{2}{123}}\right)
\qquad\text{and}\qquad
S_2\!\left(-\sqrt{\tfrac{3}{82}},\; -5\sqrt{\tfrac{3}{82}},\; -\sqrt{\tfrac{2}{123}}\right).
\end{equation*}
The positive signs give \(S_1\text{,}\) the point closest to the plane, and the negative signs give \(S_2\text{,}\) the farthest. At both points the tangent plane is parallel to the plane \(x + 5y + 2z = 3\text{.}\)
