A normal chord PQ drawn at a point P on the p | vedclass

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(C) For the parabola $y^2 = 4ax$,where $4a = 5$,so $a = \frac{5}{4}$.Let the point $P$ be $(at^2, 2at)$.The normal at $P(at^2, 2at)$ meets the parabol

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$(10, -5\sqrt{2})$

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(C) For the parabola $y^2 = 4ax$,where $4a = 5$,so $a = \frac{5}{4}$.Let the point $P$ be $(at^2, 2at)$.The normal at $P(at^2, 2at)$ meets the parabola again at $Q(at_1^2, 2at_1)$,where $t_1 = -t - \frac{2}{t}$.The chord $PQ$ subtends a right angle at the vertex $(0,0)$,so the product of slopes of $OP$ and $OQ$ is $-1$.Slope of $OP = \frac{2at}{at^2} = \frac{2}{t}$.Slope of $OQ = \frac{2at_1}{at_1^2} = \frac{2}{t_1}$.Thus,$\left(\frac{2}{t}ight) 	imes \left(\frac{2}{t_1}ight) = -1 \implies t_1 = -\frac{4}{t}$.Equating the two expressions for $t_1$: $-t - \frac{2}{t} = -\frac{4}{t} \implies t = \frac{2}{t} \implies t^2 = 2 \implies t = \sqrt{2}$ (since $P$ is in the first quadrant).Then $t_1 = -\frac{4}{\sqrt{2}} = -2\sqrt{2}$.The coordinates of $Q$ are $(at_1^2, 2at_1) = \left(\frac{5}{4}(-2\sqrt{2})^2, 2(\frac{5}{4})(-2\sqrt{2})ight) = \left(\frac{5}{4}(8), -5\sqrt{2}ight) = (10, -5\sqrt{2})$.

$A$ normal chord $PQ$ drawn at a point $P$ on the parabola $y^2 = 5x$ subtends a right angle at the vertex. If $P$ lies in the first quadrant,then the other end $Q$ of the normal chord is

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