The slope of a chord of the parabola y^2 = 4a | vedclass

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(B) Let the chord join the points $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since the chord subtends a right angle at the origin,the product of slopes

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$\sqrt{2}$

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(B) Let the chord join the points $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.Since the chord subtends a right angle at the origin,the product of slopes of $OP$ and $OQ$ is $-1$,so $\frac{2at_1}{at_1^2} 	imes \frac{2at_2}{at_2^2} = -1$,which gives $t_1 t_2 = -4$.Since the chord is normal at $P$,the slope of the normal at $P$ is $-t_1$. The slope of the chord $PQ$ is $m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2}{t_1 + t_2}$.Equating the slope of the normal to the slope of the chord: $-t_1 = \frac{2}{t_1 + t_2} \implies -t_1(t_1 + t_2) = 2 \implies -t_1^2 - t_1 t_2 = 2$.Substituting $t_1 t_2 = -4$,we get $-t_1^2 - (-4) = 2 \implies t_1^2 = 2 \implies t_1 = \pm \sqrt{2}$.The slope of the chord is $m = -t_1 = \mp \sqrt{2}$. Since the magnitude is requested,the slope is $\sqrt{2}$.

The slope of a chord of the parabola $y^2 = 4ax$ which is normal at one end and which subtends a right angle at the origin is

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