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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)

intersects the parabola

x^2 = 4y

A

and

B

. Then the locus of the centroid of

\triangle OAB

is:

3x^2 - 2x = 4y

3y^2 - 2y = 4x

3x^2 + 2x = 4y

correct

Dnone of these

Solution

Step 1: With

A(2t_1, t_1^2),\ B(2t_2, t_2^2)

and

P(-1, 0)

collinear,

\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)

gives

t_1 + t_2 = \dfrac{3h}{2}

and

t_1 t_2 = -\dfrac{3h}{4}

, so

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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