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?

  • $\begingroup$ 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

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.

enter image description here

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.