# Finding a normal to an ellipsoid

Let $E$ be an ellipsoid centered at $v = (x,y,z) \in \mathbb{R}^3$ and let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation which transforms $E$ to a sphere $S$ with a radius of length $1$. Assume $p \in \mathbb{R}^3$ is on the surface of $E$, how can I find the normal through that point?

Since T is an isomorphism, the ellipsoid consists of all points $\textbf{p}$ such that $(\textbf{p}-\textbf{v})^{T}T^{T}T(\textbf{p}-\textbf{v})=1$. Let $\textbf{x}(t)=\textbf{p}(t)-\textbf{v}$ where $\textbf{p}(t)$ is a smooth curve on the ellipsoid starting at $\textbf{p}_{0}$.
Then $\textbf{u}:=\textbf{x}'(0)$ is tangent to the ellipsoid at the point $\textbf{p}(0)$. Furthermore: $$\frac{\textbf{d}}{\textbf{d}t}(\textbf{x(t)}^{T} T^{T}T\textbf{x}(t))=2 \textbf{x}'(t)^{T}T^{T}T\textbf{x}(t)=\frac{\textbf{d}}{\textbf{d}t}(1)=0$$ for all t. So $\textbf{u}^{T} T^{T} T (\textbf{p}_{0}-\textbf{v})=0$. The tangent vector $\textbf{u}$ can be chosen arbitrarily so $\textbf{n}= T^{T} T (\textbf{p}_{0}-\textbf{v})$ is normal to the ellipsoid at point $\textbf{p}_{0}$.