I was looking at this ellipsoid:
$$
\frac{x^2}{25}+\frac{y^2}{25}+\frac{z^2}{9}=1
$$
I tried parametrizing it as such:
$$
\gamma\left(\theta, \varphi\right)=\left(5\cos\left(\theta\right)\sin\left(\varphi\right), 5\sin\left(\theta\right)\sin\left(\varphi\right), 3\cos\left(\varphi\right)\right)
$$
With $\varphi\in\left[0,\pi\right]$, $\theta\in\left[0,2\pi\right]$.
I was curious about vectors normal to its surface, which to my understanding (which goes with my intuition well, too) should include vectors of the form $\frac{\partial\gamma}{\partial\theta}\times\frac{\partial\gamma}{\partial\varphi}$.
I went for a point that I thought would be easy to deduce a normal vector out of, and then went on to calculate it, hoping to obtain a result that matches my expectations, but it failed.
Specifically, I took $\varphi=\pi$, so decided to look at the lowest point: $\left(0,0,-3\right)$. The surface there seems to be parallel to the $XY$ plane, so my assumption was that I'd obtain a normal vector parallel to it: so $c\hat{z}$ for some $0\neq c\in\mathbb{R}$.
What I got instead, however, was... the zero vector, $\vec{0}=\left(0,0,0\right)$! I'm not sure why and how that makes sense. (I'll attach my calculations below. I attempted to validate them, both manually and using WolframAlpha, where I could, but failed to find an oopsie (which most definitely does not rule out the existence of one (or more))).
If this isn't a calculation error, I'd appreciate any help with understanding the meaning of this result. Is it something to do with the parametrization I went for, that causes my two derivative vectors to be parallel by chance, or is it more than that?
My calculation:
$$
\vec{N}=\vec{R_{\varphi}}\times\vec{R_{\theta}}=\left|\begin{array}{ccc}
\hat{x} & \hat{y} & \hat{z}\\
5\cos\theta\cos\varphi & 5\sin\theta\cos\varphi & -3\sin\varphi\\
-5\sin\theta\sin\varphi & 5\cos\theta\sin\varphi & 0
\end{array}\right|
=\left(15\cos\theta\sin^{2}\varphi,15\sin\theta\sin^{2}\varphi,25\cos^{2}\theta\cos\varphi\sin\varphi+25\sin^{2}\theta\cos\varphi\sin\varphi\right)
=\left(15\cos\theta\sin^{2}\varphi,15\sin\theta\sin^{2}\varphi,25\cos\varphi\sin\varphi\right)
$$
And so substituting $\pi\rightarrow\varphi$ results in the whole thing going poof.
Use the definition of tangent space: you consider $f(x,y,z) = \frac{x^2}{25} + \frac{y^2}{25} + \frac{z^2}{9} -1$, $E = f^{-1}(\{0\})$. Now you know that the tangent space at $X_0= (x_0,y_0,z_0)$ is $$T_{X_0} = X_0+ \ker df_{X_0}$$ but $$df_X = \begin{pmatrix} \frac{2x}{25} & \frac{2y}{25} & \frac{2z}{9}\end{pmatrix}$$ which means your normal vector at $X$ is exactly $${df}_{X}^T = \begin{pmatrix} \frac{2x}{25} \\ \frac{2y}{25} \\ \frac{2z}{9}\end{pmatrix}$$ Or add $X$ to it regarding your definition...