JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Parabola

The parabola is $y^2=4x$, so $a=1$ and the focus is $(1,0)$.

Since AC is a focal chord passing through $(1,0)$, take $A=(t^2,2t)$; the other end of a focal chord has parameter $-1/t$, so $C=\left(\frac{1}{t^2},-\frac{2}{t}\right)$.

Length of a focal chord: $AC={\left(t+\frac{1}{t}\right)}^2$ (with $a=1$). Setting this equal to $\frac{25}{4}$:

${\left(t+\frac{1}{t}\right)}^2=\frac{25}{4}\Rightarrow t+\frac{1}{t}=\frac{5}{2}\Rightarrow t=2\text{or}\frac{1}{2}$

Taking $t=2$: $A=(4,4)$ and $C=\left(\frac{1}{4},-1\right)$.

Since AD is parallel to the y-axis, D lies on the parabola with the same x-coordinate as A: $D=(4,-4)$.

Since BC is parallel to the y-axis, B lies on the parabola with the same x-coordinate as C: $B=\left(\frac{1}{4},1\right)$.

So AD (length 8, at $x=4$) and BC (length 2, at $x=\frac{1}{4}$) are the parallel sides, and the height (horizontal distance between them) is $4-\frac{1}{4}=\frac{15}{4}$.

Area of trapezium $=\frac{1}{2}(AD+BC)\times \text{height}=\frac{1}{2}(8+2)\times \frac{15}{4}=\frac{75}{4}$