For any triple $p,q,v \in \mathbb{R^n}$ the path $$\alpha_{p,q,v}:[0,1] \to \mathbb{R^n},\alpha_{p,q,v}(t)=(1-t)p+tq+t(1-t)v$$
Is a parabolic arc with endpoints $p,q$. Prove that if $p\neq q$, for any $x\in \mathbb{R^n}- \lbrace p,q \rbrace$ there exists at most one vector $v$ orthogonal to $p-q$ and such that $x$ belongs to the range of $\alpha_{p,q,v}$
I try to imagine it in the plane and fix $x$, and draw the points $p,q$ and the vector $-q$ and draw the vector $p-q$ and try to identify the way of build the vector $v$ orthogonal to $p-q$ but Simply any construction come to my thoughts, I have 3 hours trying it, but any idea comes.
If someone can give me a suggestion, hint, or idea that was very helpful,
By a suitable translation we assume that $q=0$. For any $x$ (not necessarily belonging to $\alpha$) (and not parallel to $p$ as in this case clearly no solution exists) the usual suspect for $v$ would be a vector parallel to $$\frac{\|p\|^2}{\langle x,p\rangle}x-p,$$ that is $$v=k\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right)$$ for some constant $k$.
Now we force $k$ such $x$ belongs to $\alpha$ setting $$x=p(1-t)+t(1-t)k\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right).$$
Multiply by $p$ to get $$1-t=\frac{\langle x,p\rangle}{\|p\|^2}$$ and subtitute that $t$.
We want $0\leq t\leq1$, so we must have $$0\leq\frac{\langle x,p\rangle}{\|p\|^2}\leq1.$$
Now multiply out and determine the coefficient of $p$. As $p$ and $x$ are linearly independent the coefficient of $p$ must vanish, which yields to $$k=\frac{\|p\|^2}{\|p\|^2-\langle x,p\rangle},$$ if $p$ and $p-x$ aren't orthogonal (in which case there is no solution). Hence finally $$v=\frac{\|p\|^2}{\|p\|^2-\langle x,p\rangle}\cdot\left(\frac{\|p\|^2}{\langle x,p\rangle}x-p\right).$$