The length of the normal chord to the parabol | vedclass

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(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $P(t_1^2, 2t_1)$ meets the parabola again at $Q(t_2^2, 2t_2)$,where $t_2 = -t_1 - \

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$6\sqrt{3}$

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(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $P(t_1^2, 2t_1)$ meets the parabola again at $Q(t_2^2, 2t_2)$,where $t_2 = -t_1 - \frac{2}{t_1}$.If the chord $PQ$ subtends a right angle at the vertex $(0,0)$,then the product of the slopes of $OP$ and $OQ$ is $-1$.Slope of $OP = \frac{2t_1}{t_1^2} = \frac{2}{t_1}$ and slope of $OQ = \frac{2t_2}{t_2^2} = \frac{2}{t_2}$.So,$(\frac{2}{t_1}) 	imes (\frac{2}{t_2}) = -1 \Rightarrow t_1 t_2 = -4$.Substituting $t_2 = -\frac{4}{t_1}$ into the normal condition: $-t_1 - \frac{2}{t_1} = -\frac{4}{t_1}$ $\Rightarrow t_1 = \frac{2}{t_1}$ $\Rightarrow t_1^2 = 2$.Thus,$t_1 = \sqrt{2}$ and $t_2 = -2\sqrt{2}$.The coordinates are $P(2, 2\sqrt{2})$ and $Q(8, -4\sqrt{2})$.The length $PQ = \sqrt{(8-2)^2 + (-4\sqrt{2} - 2\sqrt{2})^2} = \sqrt{6^2 + (-6\sqrt{2})^2} = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3}$.

The length of the normal chord to the parabola $y^2 = 4x$,which subtends a right angle at the vertex,is

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