Let P(at^{2}, 2at), Q, and R(ar^{2}, 2ar) be | vedclass

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(D) Since $PQ$ is a focal chord with $P(at^{2}, 2at)$, the coordinates of $Q$ are $(\frac{a}{t^{2}}, \frac{-2a}{t})$.Slope of $QR = \frac{2ar - (-2a/t

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$\frac{t^{2}-1}{t}$

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(D) Since $PQ$ is a focal chord with $P(at^{2}, 2at)$, the coordinates of $Q$ are $(\frac{a}{t^{2}}, \frac{-2a}{t})$.Slope of $QR = \frac{2ar - (-2a/t)}{ar^{2} - a/t^{2}} = \frac{2a(r + 1/t)}{a(r - 1/t)(r + 1/t)} = \frac{2}{r - 1/t} = \frac{2t}{rt - 1}$.Slope of $PK = \frac{2at - 0}{at^{2} - 2a} = \frac{2at}{a(t^{2} - 2)} = \frac{2t}{t^{2} - 2}$.Since $PK \parallel QR$, their slopes are equal:$\frac{2t}{rt - 1} = \frac{2t}{t^{2} - 2}$.Assuming $t 
eq 0$, we have $rt - 1 = t^{2} - 2$.$rt = t^{2} - 1$.$r = \frac{t^{2} - 1}{t}$.

Let $P(at^{2}, 2at)$, $Q$, and $R(ar^{2}, 2ar)$ be three points on the parabola $y^{2}=4ax$. If $PQ$ is a focal chord and $PK$ is parallel to $QR$, where the coordinates of $K$ are $(2a, 0)$, then the value of $r$ is:

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