Mathematics · Parabola

JEE Main 2026 — 28 January, Evening Shift — Question 3

Let A be the focus of the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ . Let the line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ intersect the parabola at two distinct points B and C . If the centroid of the triangle ABC is $\left(\frac{7}{3}, \frac{4}{3}\right)$ , then $(\mathrm{BC})^{2}$ is equal to :

Option A:
41
Option B:
80
Correct
Option C:
89
Option D:
32

Answer: B

Step-by-step solution

Coordinates of centroid of triangle ABC are $\frac{2}{3}\left(\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+1\right)=\frac{7}{3}$

$\Rightarrow \mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}=\frac{5}{2}$

$\frac{4}{3}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\frac{4}{3} \Rightarrow \mathrm{t}_{1}+\mathrm{t}_{2}=1$

$\left(t_{1}+t_{2}\right)^{2}=t_{1}^{2}+t_{2}^{2}+2 t_{1} t_{2} \Rightarrow t_{1} t_{2}=\frac{-3}{4}$

$\left(t_{1}-t_{2}\right)^{2}=\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}=4$

$(B C)^{2}=4\left(t_{1}^{2}-t_{2}^{2}\right)^{2}+16\left(t_{1}-t_{2}\right)^{2}$

$\Rightarrow(\mathrm{BC})^{2}=80$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Considering a Line or a Point wrt a Parabola