JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let A be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ 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 $(B C)^{2}$ is equal to :

A
41
B
80
C
89
D
32

Solution

For $y^{2} = 8 x$ , $4 a = 8 \Rightarrow a = 2$ , focus $A = (2, 0)$ . Take parametric points $B = (2 t_{1}^{2}, 4 t_{1})$ and $C = (2 t_{2}^{2}, 4 t_{2})$ .

Centroid: $\frac{2 + 2 t _{1}^{2} + 2 t _{2}^{2}}{3} = \frac{7}{3} \Rightarrow t_{1}^{2} + t_{2}^{2} = \frac{5}{2}$ , and $\frac{4 ( t _{1} + t _{2} )}{3} = \frac{4}{3} \Rightarrow t_{1} + t_{2} = 1$ .

From $(t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$ : $1 = \frac{5}{2} + 2 t_{1} t_{2} \Rightarrow t_{1} t_{2} = - \frac{3}{4}$ .

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

$(B C)^{2} = (2 t_{1}^{2} - 2 t_{2}^{2})^{2} + (4 t_{1} - 4 t_{2})^{2} = 4 (t_{1}^{2} - t_{2}^{2})^{2} + 16 (t_{1} - t_{2})^{2} = 4 (t_{1} - t_{2})^{2} [(t_{1} + t_{2})^{2} + 4] = 4 \cdot 4 \cdot 5 = 80$ .

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