Conic Sections – Parabola – Area of Trapezium Involving Focal Chord – JEE Main 28 Jan 2025 Shift 1

Solution:

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

Since the diagonal AC passes through the focus $(1, 0)$, AC is a focal chord.

Let the coordinates of point A be $(at^2, 2at) = (t^2, 2t)$.

Since AC is a focal chord, the parameter of point C will be $-1/t$.

So, coordinates of C are $(\frac{a}{t^2}, -\frac{2a}{t}) = (\frac{1}{t^2}, -\frac{2}{t})$.

The length of the focal chord AC is given by $a(t + \frac{1}{t})^2$.

$$ 1 \cdot \left(t + \frac{1}{t}\right)^2 = \frac{25}{4} $$
Taking the square root:
$$ t + \frac{1}{t} = \frac{5}{2} $$
$$ 2t^2 – 5t + 2 = 0 $$
$$ (2t – 1)(t – 2) = 0 \Rightarrow t = 2 \text{ or } t = \frac{1}{2} $$
Let us take $t = 2$.

Coordinates of vertices:

$A = (2^2, 2(2)) = (4, 4)$.

$C = (\frac{1}{2^2}, -\frac{2}{2}) = (\frac{1}{4}, -1)$.

Since sides AD and BC are parallel to the y-axis and lie on the parabola:

Point D must have the same x-coordinate as A but opposite y-coordinate (due to symmetry).

$D = (4, -4)$.

Point B must have the same x-coordinate as C but opposite y-coordinate.

$B = (\frac{1}{4}, 1)$.

Now we calculate the dimensions of the trapezium:

Length of parallel side AD $= 4 – (-4) = 8$.

Length of parallel side BC $= 1 – (-1) = 2$.

Height (distance between parallel sides) $= x_A – x_B = 4 – \frac{1}{4} = \frac{15}{4}$.

Area of Trapezium $= \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$
$$ \text{Area} = \frac{1}{2} \times (8 + 2) \times \frac{15}{4} $$
$$ \text{Area} = \frac{1}{2} \times 10 \times \frac{15}{4} $$
$$ \text{Area} = 5 \times \frac{15}{4} = \frac{75}{4} $$

Ans. (1)