If a normal chord of the parabola y^2 = 4ax s | vedclass

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(B) Let the coordinates of the points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since $PQ$ is a normal chor

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$\pm \sqrt{2}$

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(B) Let the coordinates of the points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since $PQ$ is a normal chord,we have the relation $t_2 = -t_1 - \frac{2}{t_1} \quad \dots(1)$.The slopes of $OP$ and $OQ$ are $m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}$ and $m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2}$ respectively.Since $PQ$ subtends a right angle at the vertex $O(0,0)$,the product of the slopes of $OP$ and $OQ$ is $-1$.$\frac{2}{t_1} 	imes \frac{2}{t_2} = -1 \Rightarrow t_1t_2 = -4 \quad \dots(2)$.Substituting $t_2$ from $(1)$ into $(2)$,we get $t_1(-t_1 - \frac{2}{t_1}) = -4$.$-t_1^2 - 2 = -4$ $\Rightarrow t_1^2 = 2$ $\Rightarrow t_1 = \pm \sqrt{2}$.The slope of the normal at $P$ is $m = -t_1$.Therefore,the slope of the normal chord is $m = \mp \sqrt{2}$.Since the options represent the set of possible slopes,the correct choice is $\pm \sqrt{2}$.

If a normal chord of the parabola $y^2 = 4ax$ subtends a right angle at the vertex,its slope is-

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