JEE Advanced 2016 (4)

Shorter methods are always welcome. ,here is my simple approach

$\large{A(t_{1}^{2},2t_{1}),~B(t_{1}^{2},-2t_{1}),~C(t_{2}^{2},2t_{2}),~D(t_{2}^{2},-2t_{2})}$

Equation of diagonal $BC$

$\large{y-2t_{2}=\frac{2}{t_{2}-t_{1}} (x-t_{2}^{2})}$

and this passes through $(1,0)$ thus we have

$\large{\Rightarrow t_{1}t_{2}=1 \qquad ..(1)}$

And length of $BC$

$\large{\Rightarrow \sqrt{(t_{2}^{2}-t_{1}^{2})^2+4(t_{1}+t_{2})^2}=6.25}$

$\large{\Rightarrow (t_{1}+t_{2})^{2} \left( (t_{2}-t_{1})^2+4 \right)=\frac{625}{16}}$

Now $\large{ (t_{2}-t_{1})^2+4 \Rightarrow (t_{2}+t_{1})^2}$ from $(1)$

Thus $\large{(t_{2}+t_{1})^{4}=\frac{625}{16} \qquad ..(2)}$

On solving $(1)$ and $(2)$

Since $t_{1}$ is nearer to origin so $t_{1}=\frac{1}{2}$ and $t_{2}=2$ .

Now area of trapezium $\large{A =\frac{1}{2}(AB+CD)(AN)}$

$\large{\Rightarrow A=\frac{1}{2}(4t_{1}+4t_{2})(t_{2}^{2}-t_{1}^{2})}$

Plug in the values and

$\large{A=\frac{75}{4}}$

We see that $(1,0)$ is the focus of the parabola $y^{2} = 4x$ . It is known that the focal chord made from the point $(t^2,2t)$ has its other endpoint at $(\frac {1}{t^{2}} , \frac {-2}{t} )$ , and the length of the chord is given by $[ t+\frac {1}{t}]^2$ .

Now, $(t+\frac {1}{t})^2 = 6.25$ has roots $2,-2, \frac {1}{2}$ and $\frac {-1}{2}$ . So the trapezium has endpoints $(4,4), (-4,4), (\frac {1}{4}, 1), (\frac {-1}{4},1)$ . Clearly, its parallel sides have length $8,2$ and its height is $\frac {15}{4}$ . The trapezium has an area

$A= \frac {\frac {15}{4}}{2} \times (8+2) = \frac {75}{4}$ , so that $\frac {16A}{25} = 12$ .