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Mathematics · Parabola

JEE Main 2025 — 28 January, Morning Shift — Question 2

Let ABCD be a trapezium whose vertices lie on the parabola y2=4xy^{2}=4 x. Let the sides ADA D and BCB C 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 ABCD is :

  1. Option A:

    754\frac{75}{4}

    Correct
  2. Option B:

    252\frac{25}{2}

  3. Option C:

    1258\frac{125}{8}

  4. Option D:

    758\frac{75}{8}

Answer: A

Step-by-step solution

A(at12,2at1)&C(at12,−2at1)\mathrm{A}\left(\mathrm{at}_{1}{ }^{2}, 2 \mathrm{a} \mathrm{t}_{1}\right) \& \mathrm{C}\left(\frac{\mathrm{a}}{\mathrm{t}_{1}^{2}},-\frac{2 \mathrm{a}}{\mathrm{t}_{1}}\right)

Length AC=a(t1+1t1)2=254,t1+1t1=±52\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}

⇒t1=2\Rightarrow \mathrm{t}_{1}=2 or 12, A(14,1),D(14,−1),B(4,4),C(4,−4)\frac{1}{2}, \mathrm{~A}\left(\frac{1}{4}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4)

So, area of trapezium =12(8+2)(4−14)=754=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Parabola
Topic
Considering a Line or a Point wrt a Parabola