Question

Vertices of a trapezium lie on the parabola $y^2=2x$ and its diagonals pass through $(\frac{1}{2},0)$ with length of $\frac{25}{8}$ each. If area of trapezium is $A$, then $16A=$

Answer 1

Diagonals of trapezium $PQRS$ pass through focus $F\equiv (\frac{1}{2},0)$ of the parabola.

Length of focal chord is given by
$a{\sqrt{{\left(t^2-\frac{1}{t^2}\right)}^2+4{\left(t^{}+\frac{1}{t^{}}\right)}^2}}^{}$
$a{(t+\frac{1}{t})}^2=\frac{25}{8}$
${(t+\frac{1}{t})}^2=\frac{25}{4}$
$(t+\frac{1}{t})=\pm \frac{5}{2}$
$\Rightarrow \left(\frac{t^2+1}{t}\right)=\frac{5}{2}$ and $\left(\frac{t^2+1}{t}\right)=-\frac{5}{2}$
$2t^2+2=5t$ and $2t^2+2=-5t$
$\Rightarrow t=2,\frac{1}{2}$ and $t=-2,-\frac{1}{2}$
$\Rightarrow \frac{1}{2},2,-2,-\frac{1}{2}$ are parameters for points $P,Q,R,S$ respectively.
Therefore, coordinates of points are given by
$P\equiv (\frac{1}{8},\frac{1}{2}),Q\equiv (2,2),R\equiv (2,-2)$ and $S(\frac{1}{8},-\frac{1}{2})$
$\Rightarrow PS$ and $QR$ are the parallel sides of the trapezium.
Area of trapezium is given by
$A=\frac{1}{2}(1+4)(2-\frac{1}{8})$
$A=\frac{5}{2}\times \frac{15}{8}$
$A=\frac{75}{16}$