Let $A$ be the focus of the parabola $y^2 = 8x$. A line $y = mx + c$ intersects the parabola at two

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

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**Given:** Parabola $y^2 = 8x$, so $a = 2$ and focus $A = (2, 0)$. Parametric coordinates of points on the parabola: $B = (2t_1^2,\ 4t_1)$ and $C = (2t_2^2,\ 4t_2)$. **Step 1: Use centroid condition.** $$\frac{2 + 2t_1^2 + 2t_2^2}{3} = \frac{7}{3} \implies t_1^2 + t_2^2 = \frac{5}{2}$$ $$\frac{0 + 4t_1 + 4t_2}{3} = \frac{4}{3} \implies t_1 + t_2 = 1$$ **Step 2: Find $t_1 t_2$.** $$(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 \implies 1 = \frac{5}{2} + 2t_1 t_2 \implies t_1 t_2 = -\frac{3}{4}$$ **Step 3: Find $(t_1 - t_2)^2$.** $$(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1 t_2 = 1 + 3 = 4$$ **Step 4: Compute $(BC)^2$.** $$ (BC)^2 = (2t_1^2 - 2t_2^2)^2 + (4t_1 - 4t_2)^2 = 4(t_1+t_2)^2(t_1-t_2)^2 + 16(t_1-t_2)^2 $$ $$ = (t_1-t_2)^2\bigl[4(t_1+t_2)^2 + 16\bigr] = 4	imes[4	imes 1 + 16] = 4 	imes 20 = 80 $$ **Answer: (B) $80$**

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