Find the point of tangency between a plane and an ellipsoid

Question

So, it is given that -

The tangent plane to the ellipsoid $4x^2 + y^2 + 2z^2 = 16$ is $2x + y + 2z = k$.

I’m trying to find k, and the point of tangency between those two.

What I did - Assumed that the normal to the plane is parallel to the gradient of the ellipsoid, so I get $(8x, 2y, 4z) = (2, 1, 2)$. But then I get point $(x, y, z)$ which isn’t on the ellipsoid.

Why is my assumption wrong?

Answer 1

Substituting the plane equation into the ellipsoid we have

$$ 4 x^2 + (k - 2 x - 2 z)^2 + 2 z^2 = 16 $$

Solving for $x$ we have

$$ x = \frac{1}{4} \left(k-2 z\pm\sqrt{4 k z-8 z^2-k^2+32}\right) $$

but at tangency we need

$$ 4 k z-8 z^2-k^2+32 = 0 $$

solving now for $z$

$$ z = \frac{1}{4} \left(k\pm \sqrt{64-k^2}\right) $$

but at tangency $64-k^2 = 0$ then $k = \pm 8$

NOTE

At tangency we have

$$ (8x,2y,4z)=\lambda(2,1,2) $$

then

$$ x = \frac{\lambda}{4}\\ y = \frac{\lambda}{2}\\ z = \frac{\lambda}{2} $$

now after substitution into the ellipsoid equation we have the $\lambda$ value. Now with the tangency point, after substitution into the plane equation we get the $k$ value

Answer 2

You’ve set the two vectors equal to each other, which is too strong. If they’re parallel, then all you can say for sure is that one is a scalar multiple of the other, i.e., you need $(8x,2y,4z)=\lambda(2,1,2)$ for some $\lambda\ne0$ instead. Since you’re working in three dimensions, you can avoid introducing an extraneous variable by stating the condition as $(8x,2y,4z)\times(2,1,2)=0$ instead. This will give you three linear equations, two of which are independent.

Answer 3

The vector $(8x,2y,4z)$ is parallel to the vector $(2,1,2)$, not equal to it.

$8x=4z\implies z=2x, 2\times2y=8x\implies y=2x$

Substitute for $y,z$ in $4x^2+y^2+2z^2=16$

$\implies 4x^2+4x^2+8x^2=16x^2=16$

$\implies x=\pm1$

$(x,y,z)\equiv\pm(1,2,2)$

Depending on the value of $k$, you will have to select one solution.