Let ABCD be a trapezium whose vertices lie on the parabola y^2=4 x.

Question

Let ABCD be a trapezium whose vertices lie on the parabola $\mathrm{y}^2=4 \mathrm{x}$. Let the sides AD and BC 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

Accepted Answer

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

Solutions · 1