Coordinate Geometry – Parabola – JEE Main 28 Jan 2026 Shift 2

Solution:

For the parabola $y^2 = 8x$, comparing with $y^2 = 4ax$, we get $4a=8 \Rightarrow a=2$.

The focus A is $(a, 0) = (2, 0)$.

Let the coordinates of points B and C in parametric form be:
$$B(2t_1^2, 4t_1) \quad \text{and} \quad C(2t_2^2, 4t_2)$$

The centroid of $\Delta ABC$ is given by:
$$G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$$

Given centroid is $(\frac{7}{3}, \frac{4}{3})$.

Comparing x-coordinates:
$$\frac{2 + 2t_1^2 + 2t_2^2}{3} = \frac{7}{3}$$
$$2(1 + t_1^2 + t_2^2) = 7 \Rightarrow t_1^2 + t_2^2 = \frac{5}{2}$$

Comparing y-coordinates:
$$\frac{0 + 4t_1 + 4t_2}{3} = \frac{4}{3}$$
$$4(t_1 + t_2) = 4 \Rightarrow t_1 + t_2 = 1$$

We need to find the value of $t_1t_2$:
$$(t_1+t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$$
$$1^2 = \frac{5}{2} + 2t_1t_2$$
$$2t_1t_2 = 1 – 2.5 = -1.5 \Rightarrow t_1t_2 = -\frac{3}{4}$$

Now, calculating $(BC)^2$ using the distance formula:
$$(BC)^2 = (2t_1^2 – 2t_2^2)^2 + (4t_1 – 4t_2)^2$$
$$(BC)^2 = 4(t_1^2 – t_2^2)^2 + 16(t_1 – t_2)^2$$

We know $t_1^2 – t_2^2 = (t_1-t_2)(t_1+t_2)$. Since $t_1+t_2=1$:
$$(BC)^2 = 4[(t_1-t_2)(1)]^2 + 16(t_1-t_2)^2$$
$$(BC)^2 = 4(t_1-t_2)^2 + 16(t_1-t_2)^2 = 20(t_1-t_2)^2$$

We find $(t_1-t_2)^2$:
$$(t_1-t_2)^2 = (t_1+t_2)^2 – 4t_1t_2$$
$$(t_1-t_2)^2 = (1)^2 – 4(-\frac{3}{4}) = 1 + 3 = 4$$

Substitute this back:
$$(BC)^2 = 20(4) = 80$$

Ans. (2)