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

\sqrt{2}

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- \sqrt{2}

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\sqrt{4}

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\sqrt{6}

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Solution

The correct options are
A $\sqrt{2}$
B $- \sqrt{2}$
Let $A = (a {t_{1}}^{2}, 2 a t_{1})$ and $B = (a {t_{2}}^{2}, 2 a t_{2})$
Equation of the normal at $t_{1}$ , is given by $y - 2 a t_{1} = - t_{1} (x - a {t_{1}}^{2})$
Since it passes through $B$ , we have
$2 a t_{2} - 2 a t_{1} = - t_{1} (a {t_{2}}^{2} - a {t_{1}}^{2}) \Rightarrow 2 a (t_{2} - t_{1}) = - a t_{1} ({t_{2}}^{2} - {t_{1}}^{2})$
$\Rightarrow 2 = - t_{1} (t_{2} + t_{1})$ ...(1)
Since $O A$ and $O B$ are perpendicular to each other, we must have
$\frac{2}{t_{1}} \frac{2}{t_{2}} = - 1 \Rightarrow t_{1} t_{2} = - 4 \Rightarrow t_{2} = - \frac{4}{t_{1}}$
Putting this value of $t_{2}$ in (1), we get
$2 = - t_{1} (- \frac{4}{t_{1}} + t_{1}) = 4 - {t_{1}}^{2} = 2$ so $t_{1} = \pm \sqrt{2}$
Hence using (1)
Slope of $A B = \frac{2 a (t_{2} - t_{1})}{a ({t_{2}}^{2} - {t_{1}}^{2})} = \frac{2}{t_{1} + t_{2}} = - t_{1} = \pm \sqrt{2}$

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A is a point on the parabola y2=4ax. The normal at A cuts the parabola again at the point B. If AB subtends a right angle at the vertex of the parabola, find the slope of AB.

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