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The normal chord to a parabola $\mathrm{y^2=4 a x}$ at the point whose ordinate is equal to abscissa subtends an angle $\mathrm{\theta}$ at the focus, then $\mathrm{\frac{\theta}{2}=}$

Option: 1
$\frac{\pi}{6}$

Option: 2
$\frac{\pi}{4}$

Option: 3
$\frac{\pi}{3}$

Option: 4
$\frac{\pi}{2}$

Answers (1)

best_answer

Let the normal at $\mathrm{P\left(a t_1^2, 2 a t_1\right)}$ meets the parabola again at $\mathrm{Q\left(a t_2^2, 2 a t_2\right)}$ .

$\mathrm{\therefore \quad P Q}$ is a normal chord and

$\mathrm{ t_2=-t_1-\frac{2}{t_1} }$ ......(i)

$\mathrm{ Given \: \: 2 a t_1=a t_1^2 \Rightarrow t_1=2 }$

$\mathrm{ \therefore }$ From (i), $\mathrm{ t_2=-3 }$

Thus, $\mathrm{ P \equiv(4 a, 4 a) and Q \equiv(9 a,-6 a) }$

$\mathrm{ \therefore \quad \: Slope\: of \: S P=\frac{4 a-0}{4 a-a}=\frac{4}{3} }$

$\mathrm{and\: slope\: of \: S Q=\frac{-6 a-0}{9 a-a}=\frac{-3}{4}}$

$\mathrm{Since \: slope \: of\: S P \times\: slope of \: S Q=-1}$

$\mathrm{\therefore \angle P S Q=\frac{\pi}{2} i.e., P Q subtends\: a \, right \: \: angle \: at\: the \: focus\: S}$

Hence option 2 is correct.

Posted by

himanshu.meshram

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