Find all tangent planes of an ellipsoid which are parallel to $span \{[0,5,0],[1,5,-1]\}$

Question

I am having trouble with the following multivariabel calculus problem regarding gradients and surfaces:

I am told to find all tangent planes to the ellipsoid: $x^2+2y^2+3z^2=20$, which are parallel to the plane given by $v_1=[0,5,0]$ and $v_2=[1,5,-1]$. I get that by the cross product I can find the normal to the plane in question and thus find that $n=[5,0,5]$.

However, here I am having issues continuing, I want to say that $\nabla f(x,y,z)=[2x,4y,6z]$ which is a normal to the surface at the point $(x,y,z)$ and thus get that the normal to the tanget plane is $n_{tangent}=[5/2,0,5/6]$ but this is completely wrong, since according to the keys, there should be two of these tanget planes, and my result would only yield one.

So my question is really how to set this problem up, I cannot see how to find more than one solution, a hint or some help would be greatly appreciated.

Answer 1

This is a very standard Lagrange multiplier problem. $$\mathscr{L}(x,y,z,\lambda)=f(x)-\lambda g(x))$$ Here the function to be optimized is the plane, $f(x)=x+z$ and the constraint is the ellipsoid.

$$\mathscr{L}=x+z-\lambda\left(x^2+2y^2+3z^2-20\right)$$ $$\nabla\mathscr{L}=\left(\begin{array}{c}-2\;\lambda\;x+1\\-4\;\lambda\;y\\-6\;\lambda\;z+1\\-x^{2}-2\;y^{2}-3\;z^{2}+20\end{array}\right)=\left(\begin{array}{c}0\\0\\0\\0\end{array}\right)$$ The solution to these:$$ \left\{ \left\{ x = -\sqrt{15}, y = 0, z = -\frac{\sqrt{15}}{3}, \lambda = -\frac{\sqrt{15}}{30} \right\} , \left\{ x = \sqrt{15}, y = 0, z = \frac{\sqrt{15}}{3}, \lambda = \frac{\sqrt{15}}{30} \right\} \right\} $$ and of course you don't care about $\lambda$. One of the planes is $$\left(\begin{array}{c}5\\0\\5\end{array}\right)\cdot\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}-\sqrt{15}\\0\\-\frac{\sqrt{15}}{3}\end{array}\right)\left(\begin{array}{c}5\\0\\5\end{array}\right)$$ and the other one is with the two negative values changed to positive.

Answer 2

You know that a normal to the tangent plane is $\mathbf n = [5,0,5]$, but the point that you got from setting this equal to $\nabla f$ and solving for $x$, $y$ and $z$ isn’t even on the surface. The problem is that, unless you’re lucky, these two vectors probably don’t have the same length or even point in the same direction. The condition that you need to use is that $\nabla f$ is parallel to $\mathbf n$, i.e., that it’s some multiple of $\mathbf n$. Since you’re working in $\mathbb R^3$, this condition can be stated with a cross product: $$\mathbf n\times\nabla f = [20y, -10x+30z,-20y] =0.$$ Solve this equation together with the equation of the surface to find the two points of tangency, from which I expect that you can derive the equations of the two tangent planes.

You could also solve this without calculus by finding the values of $d$ for which the plane $5x+5z+d=0$ has only one intersection with the ellipsoid. This will reduce to solving for values of $d$ that make the discriminant of a quadratic equation vanish. However, there’s a direct way to obtain an equation for $d$ by using homogeneous coordinates: if $C$ is the matrix of a quadric, every tangent plane $\mathbf\pi$ satisfies the dual equation $\mathbf\pi^TC^*\mathbf\pi=0$. For this ellipsoid, $C=\operatorname{diag}(1,2,3,-20)$ and we can use its inverse for the dual conic $C^*$, so the equation to be solved is $$\begin{bmatrix}5&0&5&d\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&\frac12&0&0\\0&0&\frac13&0\\0&0&0&-\frac1{20}\end{bmatrix} \begin{bmatrix}5\\0\\5\\d\end{bmatrix} = \frac{100}3-\frac{d^2}{20} = 0.$$