A variable chord $PQ$ of the parabola $y^{2} = 4ax$ is drawn parallel to the line $y = x$. If the pa

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A variable chord $PQ$ of the parabola $y^{2} = 4ax$ is drawn parallel to the line $y = x$. If the parameter of the points $P$ and $Q$ on the parabola be $t_{1}$ and $t_{2}$ respectively, then :

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Sol. [Key Concept: A chord parallel to $y = x$ on $y^{2} = 4ax$ has $t_{1} + t_{2} = 2$; the intersection of tangents at its ends lies on $y = 2a$.]slope $= \frac{2}{t_{1} + t_{2}} = 1 \Rightarrow t_{1} + t_{2} = 2$locus of $R$ is$$\begin{matrix} & k = a\left( t_{1} + t_{2} ight) = 2a \ \Rightarrow \ y & \ = 2a \end{matrix}$$locus of $S$ is$$\begin{matrix} h = & a\left( \left( t_{1} + t_{2} ight)^{2} - t_{1}t_{2} + 2 ight) = a\left( 6 - t_{1}t_{2} ight) \ & k = - at_{1}t_{2}\left( t_{1} + t_{2} ight) = - 2at_{1}t_{2} \ 	herefore\ & 2h - k = 12a \ & 2x - y = 12a \end{matrix}$$

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