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JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Parabola

Let A B C D be a trapezium whose vertices lie on the parabola y^2=4...

Let

A B C D

be a trapezium whose vertices lie on the parabola

y^2=4 x

. Let the sides

A D

and

B C

of the trapezium be parallel to y -axis. If the diagonal AC is of length

\frac{25}{4}

and it passes through the point

(1,0)

, then the area of

A B C D

is

Options

A $\frac{75}{4}$
B $\frac{25}{2}$
C $\frac{125}{8}$
D $\frac{75}{8}$

Correct answer

A. $\frac{75}{4}$

Step-by-step solution

\begin{aligned} & \mathrm{A}\left(\mathrm{at}_1^2, 2 \mathrm{at}\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) \end{aligned}

So, area of trapezium =

\frac{1}{2}

(8+2) \left(4-\frac{1}{4}

\right)

=\frac{75}{4}

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