Let P(frac{pi}{4}), Q(frac{5 pi}{4}), R(frac{ | vedclass

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(D) The equation of the hyperbola is $x^2 - 4y^2 = 4$, which can be written as $\frac{x^2}{4} - \frac{y^2}{1} = 1$. Here, $a = 2$ and $b = 1$. The par

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$8$

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(D) The equation of the hyperbola is $x^2 - 4y^2 = 4$, which can be written as $\frac{x^2}{4} - \frac{y^2}{1} = 1$. Here, $a = 2$ and $b = 1$. The parametric coordinates are given by $(a \sec 	heta, b 	an 	heta) = (2 \sec 	heta, 	an 	heta)$. Calculating the points: $P(\frac{\pi}{4}) = (2 \sec \frac{\pi}{4}, 	an \frac{\pi}{4}) = (2 \sqrt{2}, 1)$ $Q(\frac{5 \pi}{4}) = (2 \sec \frac{5 \pi}{4}, 	an \frac{5 \pi}{4}) = (-2 \sqrt{2}, 1)$ $R(\frac{3 \pi}{4}) = (2 \sec \frac{3 \pi}{4}, 	an \frac{3 \pi}{4}) = (-2 \sqrt{2}, -1)$ $T(\frac{7 \pi}{4}) = (2 \sec \frac{7 \pi}{4}, 	an \frac{7 \pi}{4}) = (2 \sqrt{2}, -1)$ These points form a rectangle with vertices $(2 \sqrt{2}, 1), (-2 \sqrt{2}, 1), (-2 \sqrt{2}, -1),$ and $(2 \sqrt{2}, -1)$. The length of the sides are: Width $= |2 \sqrt{2} - (-2 \sqrt{2})| = 4 \sqrt{2}$ Height $= |1 - (-1)| = 2$ Area $= 	ext{width} 	imes 	ext{height} = 4 \sqrt{2} 	imes 2 = 8 \sqrt{2}$ square units. Thus, option $D$ is correct.

Let $P(\frac{\pi}{4}), Q(\frac{5 \pi}{4}), R(\frac{3 \pi}{4}), T(\frac{7 \pi}{4})$ be the points on the hyperbola $x^2-4y^2-4=0$ in the parametric form. Then the area of the quadrilateral $PQRT$ is (in square units) (in $\sqrt{2}$)

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