ParabolaHardBloom L3
Question
The normal chord of a parabola $y^{2} = 4ax$ at the point whose ordinate is equal to the abscissa, then angle subtended by normal chord at the focus is :
Options
A.$\frac{\pi}{4}$
B.$\tan^{- 1}\sqrt{2}$
C.$\tan^{- 1}2$
D.$\frac{\pi}{2}$
Solution
Sol. [Key Concept: Get $t$ from ordinate $=$ abscissa ($2at = {at}^{2}$); the normal chord's far end is $t' = - t - \frac{2}{t}$, and the angle at the focus follows from the focal radii.]
$$at_{1}^{2} = 2at_{1} \Rightarrow t_{1} = 2 $$
$$\begin{matrix} & t_{2} = - t_{1} - \frac{2}{t_{1}} = - 3 \\ & m_{PS} = \frac{4a - 0}{4a - a} = \frac{4}{3} \\ & m_{QS} = \frac{- 6a - 0}{9a - a} = - \frac{3}{4} \\ & m_{PS}m_{QS} = - 1 \Rightarrow PS\bot SQ \end{matrix}$$
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