PQ is a normal chord of the parabola y^2 = 4a | vedclass

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(C) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.Since $PQ$ is a normal chord at $P$,we have the relation $t_2 = -t_1

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equal to twice the focal distance of the point $P$

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(C) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.Since $PQ$ is a normal chord at $P$,we have the relation $t_2 = -t_1 - \frac{2}{t_1}$,which implies $t_1t_2 + t_1^2 = -2$.The slope of the line $AQ$ is $m = \frac{2at_2 - 0}{at_2^2 - 0} = \frac{2}{t_2}$.The line through $P(at_1^2, 2at_1)$ parallel to $AQ$ has the equation:$y - 2at_1 = \frac{2}{t_2}(x - at_1^2)$.To find the intersection with the $x$-axis,set $y = 0$:$-2at_1 = \frac{2}{t_2}(x - at_1^2) \Rightarrow -at_1t_2 = x - at_1^2$.$x = at_1^2 - at_1t_2$.Substituting $t_1t_2 = -2 - t_1^2$:$x = at_1^2 - a(-2 - t_1^2) = at_1^2 + 2a + at_1^2 = 2a + 2at_1^2 = 2(a + at_1^2)$.Since $A$ is the origin $(0,0)$,the length $AR = |x| = 2(a + at_1^2)$.The focal distance of point $P(at_1^2, 2at_1)$ is $a + at_1^2$.Thus,$AR = 2 	imes (	ext{focal distance of } P)$.

$PQ$ is a normal chord of the parabola $y^2 = 4ax$ at $P$,$A$ being the vertex of the parabola. Through $P$,a line is drawn parallel to $AQ$ meeting the $x$-axis in $R$. Then the length of $AR$ is:

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