Parameter Relation for Parallel Lines and Focal Chord in Parabola

To find the value of the parameter $r$, we proceed as follows:

1. Coordinates of Point $Q$: The point $P$ on the parabola $y^2 = 4ax$ is given by $P(at^2, 2at)$, where $t$ is its parameter. Since $PQ$ is a focal chord passing through the focus $(a, 0)$, the parameter of point $Q$, denoted as $t_Q$, is related to $t$ by: $t_Q = -\frac{1}{t}$ Thus, the coordinates of $Q$ are: $Q\left(\frac{a}{t^2}, -\frac{2a}{t}\right)$

2. Slope of Line $QR$: For any two points on the parabola $y^2 = 4ax$ with parameters $t_1$ and $t_2$, the slope of the chord joining them is given by: $m = \frac{2a t_2 - 2a t_1}{a t_2^2 - a t_1^2} = \frac{2}{t_1 + t_2}$ Applying this formula to the line $QR$ (with parameters $t_Q = -\frac{1}{t}$ and $r$): $m_{QR} = \frac{2}{r + t_Q} = \frac{2}{r - \frac{1}{t}} = \frac{2t}{rt - 1}$

3. Slope of Line $PK$: Given the points $P(at^2, 2at)$ and $K(2a, 0)$, the slope of the line $PK$ is: $m_{PK} = \frac{2at - 0}{at^2 - 2a} = \frac{2t}{t^2 - 2}$

4. Equating the Slopes: Since the line $QR$ is parallel to the line $PK$, their slopes must be equal: $m_{QR} = m_{PK}$ $\frac{2t}{rt - 1} = \frac{2t}{t^2 - 2}$

   Since $t \neq 0$, we can divide both sides by $2t$: $rt - 1 = t^2 - 2$ $rt = t^2 - 1$ $r = \frac{t^2 - 1}{t}$

Therefore, the value of the parameter $r$ is $\frac{t^2 - 1}{t}$, which corresponds to Option (D).