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

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Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{A}\left(\mathrm{at}_1^2, 2 \mathrm{at}\right) \& \mathrm{C}\left(\frac{\mathrm{a}}{\mathrm{t}_1^2},-\frac{2 \mathrm{a}}{\mathrm{t}_1}\right) \\ & \text { Length } \mathrm{AC}=\mathrm{a}\left(\mathrm{t}_1+\frac{1}{\mathrm{t}_1}\right)^2=\frac{25}{4}, \mathrm{t}_1+\frac{1}{\mathrm{t}_1}= \pm \frac{5}{2} \\ & \Rightarrow \mathrm{t}_1=2 \text { or } \frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4)
\end{aligned}$
So, area of trapezium $=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}$

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…

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