Let P((π)/(4)), Q ((5 π)/(4)), R((3 π)/(4)), T((7 π)/(4)) be the...

Equation of hyperbola $x^2-4 y^2-4=0\frac{x^2}{4}-\frac{y^2}{1}=$1 Then, $a=2, b=1$ Parametric point on hyperbola $=(a \sec \theta, b \tan \theta)$ Then, \mathrm{P}\left(\frac{\pi}{4}\right)= \left(2 \sec \left(\frac{\pi}{4}\right),1 \tan \left(\frac{\pi}{4}\right)\right) =(2 \sqrt{2}, 1) Similarly, Q \left(\frac{5 \pi}{4}\right)=\left(2 \sec \left(\frac{5 \pi}{4}\right), 1 \tan \left(\frac{5 \pi}{4}\right)\right)$$\begin{aligned} \text{ and } =(-2 \sqrt{2},1) \\ \text{ and } \mathrm{R}\left(\frac{3 \pi}{4}\right)=(-2 \sqrt{2},-1) \\ \text{ and } \mathrm{T}\left(\frac{7 \pi}{4}\right)=(2 \sqrt{2},-1) \end{aligned}$$ Area of quadrilateral PQRT(sq. unit) $$\begin{aligned} \text{ and } =\frac{1}{2}\left\{\left(x_1-x_3\right)\left(y_2-y_4\right)-\left(x_2-x_4\right)\left(y_1-y_3\right)\right\} \\ \text{ and } =\frac{1}{2}\{(2 \sqrt{2}+2 \sqrt{2})(1+1)-(-2 \sqrt{2}-2 \sqrt{2})(1+1)\} \\ \text{ and } =\frac{1}{2}\{4 \sqrt{2} \times2+4 \sqrt{2} \times2\} \\ \text{ and } =\frac{1}{2} \times16 \sqrt{2}=8 \sqrt{2} \end{aligned}$$So, option (d) is correct.