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Locus of Intersection of OP and Focal Perpendicular | JEE

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

Question
The straight line joining any point PP on the parabola y2=4axy^2 = 4ax to the vertex, and the perpendicular from the focus to the tangent at PP, intersect at RR. Then the equation of the locus of RR is:
Ax2+2y2−ax=0x^2 + 2y^2 - ax = 0
B2x2+y2−2ax=02x^2 + y^2 - 2ax = 0correct
C2x2+2y2−ay=02x^2 + 2y^2 - ay = 0
D2x2+y2−2ay=02x^2 + y^2 - 2ay = 0
Solution
Step 1: The tangent at P(at2,2at)P(at^2, 2at) is ty=x+at2ty = x + at^2, and the perpendicular from the focus (a,0)(a, 0) to it is tx+y=tatx + y = ta. Step 2: Line OPOP has slope 2t\dfrac{2}{t}, so y=2txy = \dfrac{2}{t}x, giving t=2xyt = \dfrac{2x}{y}. Step 3: Eliminating tt,
2xy⋅x+y=2xy⋅a⇒2x2+y2=2ax,\frac{2x}{y}\cdot x + y = \frac{2x}{y}\cdot a \Rightarrow 2x^2 + y^2 = 2ax,
 i.e. 2x2+y2−2ax=0.2x^2 + y^2 - 2ax = 0. Correct answer: (2)
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