# Fixed point of a line through two points on parabola.

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?

• What do you mean by synthetically ? – Yves Daoust Aug 27 '18 at 20:47
• No coordinate system or trigonometry, just euclidian geometry – Aqua Aug 27 '18 at 20:49

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.