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Point P Maximising |PR-PQ| on y²=4ax | JEE Parabola

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

Question
The point PP on the parabola y2=4axy^2 = 4ax for which ∣PR−PQ∣|PR - PQ| is maximum, where R=(−a,0)R = (-a, 0) and Q=(0,a)Q = (0, a), is:
A(a,2a)(a, 2a)correct
B(a,−2a)(a, -2a)
C(4a,4a)(4a, 4a)
D(4a,−4a)(4a, -4a)
Solution
Step 1: By the triangle inequality ∣PR−PQ∣≤RQ|PR - PQ| \le RQ, with equality when PP lies on line RQRQ: y=x+ay = x + a. Step 2: Intersecting with y2=4axy^2 = 4ax,
(x+a)2=4ax⇒(x−a)2=0⇒P=(a,2a).(x + a)^2 = 4ax \Rightarrow (x - a)^2 = 0 \Rightarrow P = (a, 2a).
 Correct answer: (1)
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