We have an $n$-dimensional ellipsoid described by: $$\frac{x_1^2}{a_1^2}+\dots+\frac{x_n^2}{a_n^2}=1$$ and we construct the hyperplane through any $x \in$ the ellipsoid which is tangent to the ellipsoid at $x$.
Prove that $D(x)$, the distance from this hyperplane to the origin, is: $$D(x)=\frac{1}{\sqrt{\frac{x_1^2}{a_1^4}+\dots+\frac{x_n^2}{a_n^4}}}$$
I know that the plane tangent to the ellipsoid at a point on the ellipsoid we can call $y_0=(y_1,\dots,y_n)$ can be written as: $$\frac{x_1y_1}{a_1^2}+\dots+\frac{x_ny_n}{a_n^2}=1$$
I don't see how to get from this description of the tangent plane to an expression for the distance to the origin only in terms of $x$. How do we deal with the fact that the description is at a specific point?
Edit: fixed the equation for D
Best Answer
I believe there is an error in the formula for the distance.
If $(x_1,\cdots,x_n)$ is your point on the ellipsoid, we can find a vector orthogonal to the ellipsoid by taking the gradient of the function \begin{equation} F = \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\cdots+\frac{x_n^2}{a_n^2}-1 \end{equation} so we'll have \begin{equation} \nabla F =2 \left(\frac{x_1}{a_1^2},\frac{x_2}{a_2^2},\cdots,\frac{x_n}{a_n^2}\right) \end{equation} The distance of the tangent plane from the origin can be found by finding a number $\lambda$ such that $\lambda\nabla F$ belongs to the plane. In this way you'll get that the distance is $\|\lambda\nabla F\|$.
In particular, we have that \begin{equation} y_i=2\lambda\frac{x_i}{a_i^2} \end{equation} so we should solve \begin{equation} 2\lambda\left(\frac{x_1^2}{a_1^4}+\frac{x_2^2}{a_2^4}+\cdots+\frac{x_n^2}{a_n^4}\right)-1=0 \end{equation} which yelds \begin{equation} \lambda=\frac{1}{2}\left(\frac{x_1^2}{a_1^4}+\frac{x_2^2}{a_2^4}+\cdots+\frac{x_n^2}{a_n^4}\right)^{-1} \end{equation} Now, \begin{equation} \|\lambda\nabla F\|=\lambda\|\nabla F\|=\lambda \cdot 2\left(\frac{x_1^2}{a_1^4}+\cdots+\frac{x_n^2}{a_n^4}\right)^{1/2} \end{equation} so \begin{equation} \|\lambda\nabla F\|=\frac{1}{2}\left(\frac{x_1^2}{a_1^4}+\cdots+\frac{x_n^2}{a_n^4}\right)^{-1}\cdot 2\left(\frac{x_1^2}{a_1^4}+\cdots+\frac{x_n^2}{a_n^4}\right)^{1/2}=\left(\frac{x_1^2}{a_1^4}+\cdots+\frac{x_n^2}{a_n^4}\right)^{-1/2} \end{equation}