Find tangent plane to ellipsoid $x^2+2y^2+z^2=4$ : can't find where I made a mistake...

Question

So I'm struggling on a pretty simple problem on Brilliant.org.

Find the unit normal vector to the surface of the ellipsoid defined by :

$x^2+2y^2+z^2=4$

at point : $(1;\frac{1}{\sqrt2};\sqrt2)$

What I did :

I considered the ellipsoid could be described like a rotation around the y axis of the function

$f(y)=x=2\sqrt{1-\frac{y^2}{2}}$

So to parametrize the surface I used :

$u=y$ ; $ and$ $\overrightarrow{x}(u,v) = (2\sqrt{1-\frac{u^2}{2}}\cos{v};$ $ u$ $;$ $2\sqrt{1-\frac{u^2}{2}}\sin{v})$

And derived

$\nabla\overrightarrow{x}(u,v)=((\frac{-2u}{\sqrt{1-\frac{y^2}{2}}}\cos{v};$ $ 1$ $;$ $\frac{-2u}{\sqrt{1-\frac{y^2}{2}}}\sin{v})$ , $(-2\sqrt{1-\frac{u^2}{2}}\sin{v};$ $ 0$ $;$ $2\sqrt{1-\frac{u^2}{2}}\cos{v}))$

at point : $(1;\frac{1}{\sqrt2};\sqrt2)$

$u=y=\frac{1}{\sqrt2}$ and $\cos{v}=\frac{1}{\sqrt{3}}$

$\sin{v}=\frac{\sqrt{2}}{\sqrt{3}}$

So : $\nabla\overrightarrow{x}(u,v)=((\frac{-2\sqrt{2}}{3};$ $ 1$ $;$ $\frac{-4}{3})$ , $(-\sqrt{2}$ ; $0$ ; $1))$

I computed the cross product $\overrightarrow{c}$ of the two components $\overrightarrow{a}$ and $\overrightarrow{b}$ of $\nabla\overrightarrow{x}$ ; and divided by its magnitude to have a unit vector.

$c_{x}=1/\sqrt{11}$

$c_{y}=2\sqrt{2}/\sqrt{11}$

$c_{z}=\sqrt{2}/\sqrt{11}$

Well obviously I was wrong, but the problem is, the correction uses spherical coordinates, so it doesn't help me see where I screwed up.

Their solution was :

$c_{x}=1/\sqrt{5}$

$c_{y}=\sqrt{2}/\sqrt{5}$

$c_{z}=\sqrt{2}/\sqrt{5}$

If anyone can tell me where is my error?

Thank you all very much.

Answer 1

I think you could do it much easier:

Let $F(x,y,z)= x^2 + 2y^2 + z^2 - 4 = 0$.

The gradient is always perpendicular to the surface, so $\nabla F(x,y,z) = (2x, 4y, 2z)$

$\nabla F(1,\frac{1}{\sqrt2},\sqrt2) = (2\sqrt2, \frac{4}{\sqrt2}, 2\sqrt2)$

Normalize this vector by dividing it by its own length. This yields $\frac{\nabla F(1,\frac{1}{\sqrt2},\sqrt2)}{\vert \vert \nabla F(1,\frac{1}{\sqrt2},\sqrt2) \vert \vert}$ $= (\frac{1}{\sqrt5}, \frac{\sqrt2}{\sqrt5}, \frac{\sqrt2}{\sqrt5})$ .

Answer 2

You have a stray factor of $2$ in your $u$-derivatives. It looks like you might’ve neglected the factor of $1/2$ under the radical, i.e., you differentiated $1-u^2$ instead of $1-u^2/2$.

This seems like way too much work to go through in order to find a tangent plane to this particular surface, but I suppose that the point of the exercise is to practice computing surface normals from a parameterization. You could simply compute the gradient $\nabla(x^2+2y^2+z^2)$ instead to get a surface normal. As well, if all that you’re after is an equation of the tangent plane, I don’t see a compelling reason to normalize this surface normal.