question_answer

A trapezium is inscribed in the parabola \[{{y}^{2}}=4x\] such that its diagonal pass through the point \[(1,0)\] and each has length\[\frac{25}{4}\]. Then the area of trapezium is equal to (in sq. units)

A) \[75/4\]

B) \[73/4\]

C) \[75/2\]

D) \[70/3\]

Correct Answer: A

Solution :

Focus of the parabola \[{{y}^{2}}=4x\,\operatorname{is}\,\left( 1,0 \right)\]so diagonals are focal chord

\[AS=1+{{t}^{2}}=c\] (say)

\[\because \]\[\frac{1}{c}+\frac{1}{\frac{25}{4}-c}=1\]   \[\left[ \because \frac{1}{AS}+\frac{1}{CS}=\frac{1}{a} \right]\]

\[\frac{25}{4}=\frac{25}{4}c-{{c}^{2}}\]\[\Rightarrow \]\[4{{c}^{2}}-25c+25=0\Rightarrow c=\frac{5}{4},5\]

For \[c=\frac{5}{4},1+{{t}^{2}}=\frac{5}{4}\]\[\Rightarrow \]\[{{t}^{2}}=\frac{1}{4}\Rightarrow t=\pm \frac{1}{2}\]

For \[c=5,1+{{t}^{2}}=5\Rightarrow t=\pm 2\]

\[A\equiv \left( \frac{1}{4},1 \right),B\equiv (4,4),C\equiv (4,4)\]and \[D\equiv \left( \frac{1}{4},-1 \right)\]

\[AD=2\]and\[BC=8,\]Distance between AD and \[BC=\frac{15}{4}\]

\[\therefore \]Area of trapezium \[ABCD\]\[=\frac{1}{2}\left( 2+8 \right)\times \frac{15}{4}=\frac{75}{4}\operatorname{sq}.\,units.\]