Find the slope of the normal chord of the parabola $y^{2} = 4 a x$ which subtends a right angle at its vertex.

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Solution

$\begin{matrix} L e t e n d o f c h o r d i s (a t_{1}^{2}, 2 a t_{1}) a n d (a t_{2}^{2}, 2 a t_{2}) \Rightarrow sin c e, n o r m a l t o t h e p a r a b o l a s o, t_{2} = (- t, \frac{- 2}{t_{1}}) \to (i) \Rightarrow e q u a t i o n o f n o r m a l t o t h e p a r a b o l a, y = - t x = 2 a t + a t^{3} \Rightarrow c h o r d m a k e r i g h t a n g l e a t o r i g i n \Rightarrow [\frac{(2 a t_{1} - 0)}{(a t_{1}^{2} - 0)}] [\frac{(2 a t_{1} - 0)}{a t_{1}^{2} - 0}] = - 1 \Rightarrow∴ (t_{1} t_{2}) = - 4 \Rightarrow t_{1} (- t_{1} \frac{- 2}{t_{1}}) = - 4 \Rightarrow t_{1} = - \sqrt{2} ∴ a n g l e = {tan}^{- 1} \sqrt{2} A n s . \end{matrix}$

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