JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Parabola

Let A be the focus of the parabola

y^{2} = 8 x

. Let the line

y = mx + c

intersect the parabola at two distinct points B and C . If the centroid of the triangle ABC is

(\frac{7}{3}, \frac{4}{3})

, then

(BC)^{2}

is equal to :

Answer: (B)

Solution

For

y^{2} = 8 x

4 a = 8

a = 2

and the focus is

A = (2, 0)

. Take the chord endpoints as

B = (2 t_{1}^{2}, 4 t_{1})

and

C = (2 t_{2}^{2}, 4 t_{2})

. The centroid conditions give

\frac{4 ( t _{1} + t _{2} )}{3} = \frac{4}{3}

, so

t_{1} + t_{2} = 1

, and

\frac{2 ( t _{1}^{2} + t _{2}^{2} ) + 2}{3} = \frac{7}{3}

, so

t_{1}^{2} + t_{2}^{2} = \frac{5}{2}

. Then

2 t_{1} t_{2} = (t_{1} + t_{2})^{2} - (t_{1}^{2} + t_{2}^{2}) = 1 - \frac{5}{2} = - \frac{3}{2}

, and

(t_{1} - t_{2})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2} = 1 + 3 = 4

. Therefore

(B C)^{2} = (2 t_{1}^{2} - 2 t_{2}^{2})^{2} + (4 t_{1} - 4 t_{2})^{2} = 4 (t_{1} + t_{2})^{2} (t_{1} - t_{2})^{2} + 16 (t_{1} - t_{2})^{2} = 4 (1) (4) + 16 (4) = 80.

Hence

(B C)^{2} = 80

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option