$A$ is a point on $y^{2} = 4 a x$ . The normal at $A$ meets the parabola again at $B$ . If $A B$ subtends a right angle at its vertex then the slope of $A B$ is

\frac{1}{2}

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\frac{- 1}{2}

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\sqrt{2}

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\frac{2}{3}

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Solution

The correct option is D $\sqrt{2}$
$y + t_{1} x = 2 a t_{1} + a {t_{1}}^{3}$ ( $e q^{n}$ at normal at A)
passage through B
$(2 a t_{2}) + t_{1} a {t_{2}}^{2} = 2 a t_{1} + a {t_{1}}^{3}$
On simplifying we get
$t_{2} = - t_{1} (\frac{- 2}{t_{1}})$
slope $O A$ = $m_{1} = \frac{2 a t_{1}}{a {t_{1}}^{2}} = \frac{2}{t_{1}}$
slope of $O B$ = $m_{2} = \frac{2 a t_{2}}{a {t_{2}}^{2}} = \frac{2}{t_{2}}$
$m_{1} m_{2} = - 1$
$t_{1} t_{2} = - 4$
$t_{1} t_{2} = - {t_{1}}^{2} - 2$
${t_{1}}^{2} = 2$
$t_{1} = \pm \sqrt{2}$
slope of normal $= - t_{1} =$ slope of AB
slope $= \pm \sqrt{2}$

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