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Line Through y²=4ax and x²=4ay Intersections | JEE Parabola

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

Question
If a≠0a \neq 0 and the line 2bx+3cy+4d=02bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2=4axy^2 = 4ax and x2=4ayx^2 = 4ay, then:
Ad2+(2b+3c)2=0d^2 + (2b + 3c)^2 = 0correct
Bd2+(3b−2c)2=0d^2 + (3b - 2c)^2 = 0
Cd2+(2b−3c)2=0d^2 + (2b - 3c)^2 = 0
Dd2+(3b+2c)2=0d^2 + (3b + 2c)^2 = 0
Solution
Step 1: Substituting y=x24ay = \dfrac{x^2}{4a} into y2=4axy^2 = 4ax gives x(x3−64a3)=0x(x^3 - 64a^3) = 0, so the parabolas meet at (0,0)(0, 0) and (4a,4a)(4a, 4a). Step 2: The line passes through both points.
(0,0): 4d=0⇒d=0;(4a,4a): 4a(2b+3c)+4d=0⇒2b+3c=0.(0,0):\ 4d = 0 \Rightarrow d = 0; \qquad (4a, 4a):\ 4a(2b + 3c) + 4d = 0 \Rightarrow 2b + 3c = 0.
 Step 3: Hence d2+(2b+3c)2=0d^2 + (2b + 3c)^2 = 0. Correct answer: (1)
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