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If $V$ is a vertex (actually it can be any point on parabola) of parabola $\mathcal{P}$ and $A$ and $B$ are such variable points on $\mathcal{P}$ that $\angle AVB = 90^{\circ}$. Prove that the line $AB$ goes through a fixed point.


I can prove that analyticaly:

Say $y^2 = 4p^2x$, then $V(0,0)$, $A(a^2,2pa)$ and $B(b^2,2pb)$ for some real $a$ and $b$ with condition $4p^2 = -ab$ (because $\angle AVB = 90^{\circ}$). Now the line $AB$ is $$y ={2p\over a+b}x +{2pab\over a+b}$$ which intersect the $x$-axis at $x_0=-ab = 4p^2$, which is clearlyconstant.

But how to prove it syntheticaly?

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  • $\begingroup$ What do you mean by synthetically ? $\endgroup$ – Yves Daoust Aug 27 '18 at 20:47
  • $\begingroup$ No coordinate system or trigonometry, just euclidian geometry $\endgroup$ – Aqua Aug 27 '18 at 20:49
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Let $S$ be the focus, $T$ the intersection between line $AB$ and the axis, $H$ and $K$ the projections of $A$ and $B$ on the axis. We want to prove that $VT$ is of fixed length.

We'll repeatedly use Apollonius' definition of parabola, which entails: $$ AH^2=4VS\cdot VH,\quad BK^2=4VS\cdot VK. $$ From the similitude of triangles $AHT$, $BKT$ we have $TK:TH=BK:AH$, and $(TK+TH):TH=(BK+AH):AH$, hence (supposing WLOG that $VK>VH$): $$ \begin{align} TH &={TK+TH\over BK+AH}AH={VK-VH\over BK+AH}AH={1\over4VS}{BK^2-AH^2\over BK+AH}AH\\ &={BK-AH\over4VS}AH={BK\cdot AH\over4VS}-{AH^2\over4VS}={BK\cdot AH\over4VS}-VH.\\ \end{align} $$ It follows that: $\displaystyle VT=TH+VH={BK\cdot AH\over4VS}$.

But from Pythagoras' theorem we also have: $$ \begin{align} AB^2 &=(AH+BK)^2+(VK-VH)^2=AH^2+VH^2+BK^2+VK^2+2AH\cdot BK-2VH\cdot VK\\ &=AV^2+BV^2+2AH\cdot BK-2VH\cdot VK=AB^2+2AH\cdot BK-2VH\cdot VK. \end{align} $$ It follows that: $$ AH\cdot BK=VH\cdot VK={1\over16VS^2}AH^2\cdot BK^2, \quad\text{whence:}\quad AH\cdot BK=16VS^2. $$ Substituting that into our previous result for $VT$ gives then $VT=4VS$, and the proof is complete.

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