JEE Main 2026 — 5 April, Evening Shift — Question 36

Given parabola: $y^2 = 8x$. Directrix: $x = -2$, so $A(-2,0)$. Let $B(2t^2, 4t)$ on parabola. Slope of AB: $\frac{4t}{2t^2+2} = \frac{3}{5}$. Cross-multiply: $20t = 6t^2 + 6$. $3t^2 - 10t + 3 = 0$. $t = 3$ or $t = \frac{1}{3}$. Since $\alpha > 1$, $t = 3$. Thus $B(18,12)$. Focal chord BC: other end $C \left( \frac{2}{t^2}, -\frac{4}{t} \right) = \left( \frac{2}{9}, -\frac{4}{3} \right)$. Area of $\triangle ABC$: $\frac{1}{2} \left| -2(12 + \frac{4}{3}) + 18(-\frac{4}{3} - 0) + \frac{2}{9}(0 - 12) \right|$. Simplify: $\frac{1}{2} \left| -2 \cdot \frac{40}{3} + 18 \cdot \left(-\frac{4}{3}\right) + \frac{2}{9} \cdot (-12) \right| = \frac{1}{2} \left| -\frac{80}{3} - 24 - \frac{8}{3} \right| = \frac{1}{2} \cdot \frac{160}{3} = \frac{80}{3}$. Six times area: $6 \times \frac{80}{3} = 160$.