Question

Let ABCD be a trapezium whose vertices lie on the parabola \(y^{2}=4 x.\) 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 ABCD is :

(1) \(\frac{75}{4}\)

(2) \(\frac{25}{2}\)

(3) \(\frac{125}{8}\)

(4) \(\frac{75}{8}\)

Answer 1

Correct option is (1) \(\frac{75}{4}\)   

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

Length \(\mathrm{AC}=\mathrm{a}\left(\mathrm{t}_{1}+\frac{1}{\mathrm{t}_{1}}\right)^{2}=\frac{25}{4}, \mathrm{t}_{1}+\frac{1}{\mathrm{t}_{1}}= \pm \frac{5}{2}\)

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

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