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Find a if Normals Meet on y²=4ax for Chord lx+my=1 | JEE

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

Question
Let the line lx+my=1lx + my = 1 intersect the parabola y2=4axy^2 = 4ax at points PP and QQ. The normals at PP and QQ meet at the point RR. If RR lies on the parabola y2=4axy^2 = 4ax, then aa equals:
A1l\dfrac{1}{l}
B12l\dfrac{1}{2l}
C−1l-\dfrac{1}{l}
D−12l-\dfrac{1}{2l}correct
Solution
Step 1: A point (at2,2at)(at^2, 2at) on lx+my=1lx + my = 1 gives alt2+2amt−1=0alt^2 + 2amt - 1 = 0, so the parameters of P,QP, Q satisfy t1t2=−1alt_1 t_2 = -\dfrac{1}{al}. Step 2: Normals at t1,t2t_1, t_2 meet on the parabola only if t1t2=2t_1 t_2 = 2. Therefore
−1al=2⇒a=−12l.-\frac{1}{al} = 2 \Rightarrow a = -\frac{1}{2l}.
 Correct answer: (4)
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