Let $a \in \mathbb{R}^n$ be such that $\| a \| <1$. Find a maximum of the function $$\{ x \in \mathbb{R}^n \: : \: \| x \| \le1 \} =\mathbb{B}^n \ni x \mapsto \frac{\| x-a \|^2}{(1-\langle a,x \rangle)^2} \in \mathbb{R}.$$
One can prove (using necessary condition for having extrema and writing $x$ in some orthogonal basis $a,b_1,...,b_{n-1}$) that it is achieved for boundary $x=a+b$ (with norm $1$) and $b \in \text{span}\{b_1,...,b_{n-1}\}$. The maximum is then equal to $\frac{1}{(1-\| a\|^2)^2}$. Since the computations are quite long (and I suspect that there is a simpler solution) I would like to ask You for help with finding one.
Write $x=\lambda a + b$ with $b \bot a$, with $\|x\|^2 = \lambda^2 \|a\|^2 + \|b\|^2 \le 1 $.
${\|x-a\|^2 \over (1-\langle x , a \rangle )^2} = { (1-\lambda)^2 \|a\|^2+ \|b\|^2\over (1-\lambda \|a\|^2)^2} $.
It is clear that if $\|x\|^2 <1$ we can increase $\|b\|$, hence at a $\max$ we have $\lambda^2 \|a\|^2 + \|b\|^2 = 1 $.
Hence the problem is to choose $\lambda$ with $|\lambda| \le {1 \over \|a\|}$, to maximise $c(\lambda)={ (1-\lambda)^2 \|a\|^2+ 1-\lambda^2 \|a\|^2 \over (1-\lambda \|a\|^2)^2} = { 1+ \|a\|^2 -2 \lambda \|a\|^2 \over (1-\lambda \|a\|^2)^2}$.
Since $c'(\lambda) = {2 \|a\|^4 (\lambda-1) \over (1-\lambda \|a\|^2)^3}$, we see that $c$ is maximised with $\lambda = 1$ with value $c(1) = {1 \over 1-\|a\|^2 }$.
Hence the solution is $x=a+b$ where $b \bot a$ and $\|b\|^2 = 1- \|a\|^2$.