JEE Main 2026 — Parabola Question with Solution

For the parabola $P:y^2=8x$, we have $4a=8\Rightarrow a=2$.

The equation of the directrix is $x=-a\Rightarrow x=-2$. Since the directrix cuts the $x$-axis at $A$, the coordinates of $A$ are $(-2,0)$.

Let the coordinates of point $B$ on the parabola be $(2t_1^2,4t_1)$.
The slope of $AB$ is given as $\frac{3}{5}$.

$\frac{4t_1-0}{2t_1^2-(-2)}=\frac{3}{5}$

$\frac{2t_1}{t_1^2+1}=\frac{3}{5}$

$10t_1=3t_1^2+3\Rightarrow 3t_1^2-10t_1+3=0$

$(3t_1-1)(t_1-3)=0\Rightarrow t_1=\frac{1}{3}$ or $t_1=3$.

For $t_1=\frac{1}{3}$, $\alpha =2{\left(\frac{1}{3}\right)}^2=\frac{2}{9}$, which is rejected since $\alpha >1$.

For $t_1=3$, $\alpha =2(3{)}^2=18>1$. Thus, $B$ is $(18,12)$.

Since $BC$ is a focal chord, the parameter for $C$ is $t_2=-\frac{1}{t_1}=-\frac{1}{3}$.

The coordinates of $C$ are $\left(2{\left(-\frac{1}{3}\right)}^2,4\left(-\frac{1}{3}\right)\right)=\left(\frac{2}{9},-\frac{4}{3}\right)$.

The area of $△ABC$ is given by:
$\Delta =\frac{1}{2}\left|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\right|$

$\Delta =\frac{1}{2}\left|-2\left(12-\left(-\frac{4}{3}\right)\right)+18\left(-\frac{4}{3}-0\right)+\frac{2}{9}(0-12)\right|$

$\Delta =\frac{1}{2}\left|-2\left(\frac{40}{3}\right)-24-\frac{24}{9}\right|$

$\Delta =\frac{1}{2}\left|-\frac{80}{3}-24-\frac{8}{3}\right|=\frac{1}{2}\left|-\frac{88}{3}-\frac{72}{3}\right|=\frac{1}{2}\left|-\frac{160}{3}\right|=\frac{80}{3}$

Six times the area of $△ABC=6\times \frac{80}{3}=160$.

Answer: $160$