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Minimum Projection of Latus Rectum on a Tangent | JEE

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

Question
If BCBC is a latus rectum of the parabola y2=4axy^2 = 4ax and AA is the vertex, then the minimum length of the projection of BCBC on a tangent drawn in the portion BACBAC is:
A2 a\sqrt{2}\,a
B22 a2\sqrt{2}\,acorrect
C2a2a
D32 a3\sqrt{2}\,a
Solution
Step 1: BCBC is vertical of length 4a4a; a tangent at (at2,2at)(at^2, 2at) makes angle θ\theta with tan⁡θ=1t\tan\theta = \dfrac{1}{t}, so its projection of BCBC is
4asin⁡θ=4a1+t2.4a\sin\theta = \frac{4a}{\sqrt{1 + t^2}}.
 Step 2: Over the portion BACBAC, −1≤t≤1-1 \le t \le 1, and the projection is least at t=±1t = \pm 1:
4a2=22 a.\frac{4a}{\sqrt{2}} = 2\sqrt{2}\,a.
 Correct answer: (2)
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