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Locus of Centroid of Triangle OAB on x²=4y | JEE Parabola

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

Question
A line from (−1,0)(-1, 0) intersects the parabola x2=4yx^2 = 4y at AA and BB. Then the locus of the centroid of △OAB\triangle OAB is:
A3x2−2x=4y3x^2 - 2x = 4y
B3y2−2y=4x3y^2 - 2y = 4x
C3x2+2x=4y3x^2 + 2x = 4ycorrect
Dnone of these
Solution
Step 1: With A(2t1,t12), B(2t2,t22)A(2t_1, t_1^2),\ B(2t_2, t_2^2) and P(−1,0)P(-1, 0) collinear,
t1+t22=t222t2+1⇒t1+t2=−2t1t2.\frac{t_1 + t_2}{2} = \frac{t_2^2}{2t_2 + 1} \Rightarrow t_1 + t_2 = -2t_1 t_2.
 Step 2: The centroid (h,k)(h, k) gives t1+t2=3h2t_1 + t_2 = \dfrac{3h}{2} and t1t2=−3h4t_1 t_2 = -\dfrac{3h}{4}, so
3k=t12+t22=9h24+3h2⇒3x2+2x=4y.3k = t_1^2 + t_2^2 = \frac{9h^2}{4} + \frac{3h}{2} \Rightarrow 3x^2 + 2x = 4y.
 Correct answer: (3)
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