Verified Solution • 2025–26 Session moderate

QJEEM25Ma342d: JEE Main 2025 Mathematics - Class 11 JEEM25Ma342d Solution: Let ABCD be a trapezium whose vertices lie on the parabola .…

Let ABCD be a trapezium whose vertices lie on the parabola y2=4x\mathrm{y}^2=4 \mathrm{x} . Let the sides AD and BC of the trapezium be parallel to yy -axis. If the diagonal AC is of length 254\frac{25}{4} and it passes through the point (1,0)(1,0) , then the area of ABCDA B C D is

Multiple Choice Options

  • A 758\frac{75}{8}
  • B 1258\frac{125}{8}
  • C 252\frac{25}{2}
  • D 754\frac{75}{4}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 28th January Morning Shift Mathematics - Parabola Question 1 English Explanation

amp;A(at12,2at1)&C(at12,−2at1)amp; Length AC=a(t1+1t1)2=254,t1+1t1=±52amp;⇒t1=2 or 12, A(12,1),D(14,−1),B(4,4),C(4,−4)amp; So, area of trapezium =12(8+2)(4−14)=754\begin{aligned}& \mathrm{A}\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \mathrm{C}\left(\frac{\mathrm{a}}{\mathrm{t}_1^2},-\frac{2 \mathrm{a}}{\mathrm{t}_1}\right) \\& \text { Length } \mathrm{AC}=\mathrm{a}\left(\mathrm{t}_1+\frac{1}{\mathrm{t}_1}\right)^2=\frac{25}{4}, \mathrm{t}_1+\frac{1}{\mathrm{t}_1}= \pm \frac{5}{2} \\& \Rightarrow \mathrm{t}_1=2 \text { or } \frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4) \\& \text { So, area of trapezium }=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}\end{aligned}

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