If two distinct chords of a parabola $x^{2} = 4 a y$ passing through $(2 a, a)$ are bisected on the line $x + y = 1$ , then length of latus rectum can be

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Solution

The correct options are
A $2$
B $1$
Any point on the line $x + y = 1$ can betaken as $(t, 1 - t)$

Equation of chord with this as midpoint is

$y (1 - t) - 2 a (x + t) = (1 - t)^{2} - 4 a t$

It passes through $(a, 2 a)$

therefore, $t^{2} - 2 t + 2 a^{2} - 2 a + 1 = 0$

This should have two distinct real roots so

$- a < 0$

$0 < a < 1$

length of latus rectum $< 4$
so answer is $1, 2$

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