Two distinct chords of the parabola $y^{2} = 4 a x$ passing through $P (a, 2 a)$ are bisected by the line $x - y + 1 = 0$ . The possible length of the latus rectum of the parabola when $a > 0$ is

2

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5

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Solution

The correct option is D $5$

Let $A : (a t_{1}^{2}, 2 a t_{1})$ and $B : (a t_{2}^{2}, 2 a t_{2})$
Coordinates of midpoint of $P A$ is $(\frac{a + a t_{1}^{2}}{2}, \frac{2 a + 2 a t_{1}}{2})$
Coordinates of midpoint of $P B$ is $(\frac{a + a t_{2}^{2}}{2}, \frac{2 a + 2 a t_{2}}{2})$

Midpoints are on the line $x - y + 1 = 0$
$\Rightarrow a t_{1}^{2} - 2 a t_{1} + 2 - a = 0$
and $a t_{2}^{2} - 2 a t_{2} + 2 - a = 0$

To generalise it, we can say that
$a t^{2} - 2 a t + 2 - a = 0$
Since, $A$ and $B$ are distinct points on the parabola,
$D = 4 a^{2} - 4 a (2 - a) > 0 \Rightarrow a > 1$

Therefore, the length of the latus rectum $= | 4 a | > 4$

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