If the chord joining $P (a {t_{1}}^{2}, 2 a t_{1})$ and is the focal chord, then $t_{1} t_{2}$ is

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Detailed Solution

Let $y^{2} = 4 a x$ be a parabola, if $P Q$ be a focal chord.

Consider any two points on the parabola $(a t_{1}^{2}, 2 a t_{1})$ $(a t_{2}^{2}, 2 a t_{2})$

Since PQ passes through the focus $S (a, 0), so P, S, Q$

are collinear.

Thus, $s l o p e o f (P S) = s l o p e o f (Q S)$

$\begin{array}{l} \Rightarrow \frac{2 a t_{1} - 0}{a t_{1}^{2} - a} = \frac{0 - 2 a t_{2}}{a - a t_{2}^{2}} \\ \Rightarrow \frac{2 a t_{1}}{a t_{1}^{2} - a} = \frac{2 a t_{2}}{a t_{2}^{2} - a} \\ \Rightarrow \frac{2 t_{1}}{t_{1}^{2} - 1} = \frac{2 t_{2}}{t_{2}^{2} - 1} \\ \Rightarrow \frac{t_{1}}{t_{1}^{2} - 1} = \frac{t_{2}}{t_{2}^{2} - 1} \\ \Rightarrow t_{1} (t_{2}^{2} - 1) = t_{2} (t_{1}^{2} - 1) \\ \Rightarrow t_{1} t_{2} (t_{2} - t_{1}) + (t_{2} - t_{1}) = 0 \end{array}$