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Area of Trapezium on y²=4x with Focal Chord Diagonals | JEE

JEE Maths question with a full step-by-step solution.

Question
The area of a trapezium whose vertices lie on the parabola y2=4xy^2 = 4x, with its diagonals passing through (1,0)(1, 0) and having length 254\dfrac{25}{4} units each, is:
A754\dfrac{75}{4} sq. unitscorrect
B62516\dfrac{625}{16} sq. units
C254\dfrac{25}{4} sq. units
D258\dfrac{25}{8} sq. units
Solution
Step 1: For y2=4xy^2 = 4x the focus is (1,0)(1, 0), so each diagonal is a focal chord of length 254\dfrac{25}{4}. The focal radii CC and 254−C\dfrac{25}{4} - C satisfy 1r1+1r2=1a=1\dfrac{1}{r_1} + \dfrac{1}{r_2} = \dfrac{1}{a} = 1:
4C2−25C+25=0⇒C=54 or 5.4C^2 - 25C + 25 = 0 \Rightarrow C = \frac{5}{4}\ \text{or}\ 5.
 Step 2: A focal radius is 1+t21 + t^2, giving t=±12t = \pm\dfrac{1}{2} and t=±2t = \pm 2, hence
A(14,1),D(14,−1),B(4,4),C(4,−4).A\left(\frac{1}{4}, 1\right),\quad D\left(\frac{1}{4}, -1\right),\quad B(4, 4),\quad C(4, -4).
 Step 3: AD=2AD = 2 and BC=8BC = 8 are the parallel sides, a distance 4−14=1544 - \dfrac{1}{4} = \dfrac{15}{4} apart.
Area=12(2+8)⋅154=754 sq. units.\text{Area} = \frac{1}{2}(2 + 8)\cdot\frac{15}{4} = \frac{75}{4}\ \text{sq. units}.
 Correct answer: (1)
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