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Latus Rectum from Two Chords Bisected by x+y=1 | JEE

JEE Maths question with a full step-by-step solution.

Question
Two distinct chords of the parabola y2=4axy^2 = 4ax passing through (a,2a)(a, 2a) are bisected by the line x+y=1x + y = 1. The length of the latus rectum of the parabola can be:
A99
B33correct
C44
D55
Solution
Step 1: A midpoint (t,1−t)(t, 1 - t) on x+y=1x + y = 1 gives the chord y(1−t)−2a(x+t)=(1−t)2−4aty(1 - t) - 2a(x + t) = (1 - t)^2 - 4at. Through (a,2a)(a, 2a),
t2−2t+(2a2−2a+1)=0.t^2 - 2t + (2a^2 - 2a + 1) = 0.
 Step 2: Two distinct chords need a positive discriminant.
2a−2a2>0⇒0<a<1⇒0<4a<4.2a - 2a^2 > 0 \Rightarrow 0 < a < 1 \Rightarrow 0 < 4a < 4.
 Only 33 lies in (0,4)(0, 4). Correct answer: (2)
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