Let the points P_1left(frac{pi}{4}right), P_2 | vedclass

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(A) Given the hyperbola equation $\frac{x^2}{9}-\frac{y^2}{16}=1$,we have $a=3$ and $b=4$.The parametric coordinates are given by $x=a \sec 	heta$ an

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a rectangle

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(A) Given the hyperbola equation $\frac{x^2}{9}-\frac{y^2}{16}=1$,we have $a=3$ and $b=4$.The parametric coordinates are given by $x=a \sec 	heta$ and $y=b 	an 	heta$,so $x=3 \sec 	heta$ and $y=4 	an 	heta$.Calculating the coordinates for each point:$P_1\left(\frac{\pi}{4}ight) = (3 \sec \frac{\pi}{4}, 4 	an \frac{\pi}{4}) = (3\sqrt{2}, 4)$$P_2\left(\frac{3\pi}{4}ight) = (3 \sec \frac{3\pi}{4}, 4 	an \frac{3\pi}{4}) = (-3\sqrt{2}, -4)$$P_3\left(\frac{5\pi}{4}ight) = (3 \sec \frac{5\pi}{4}, 4 	an \frac{5\pi}{4}) = (-3\sqrt{2}, 4)$$P_4\left(\frac{7\pi}{4}ight) = (3 \sec \frac{7\pi}{4}, 4 	an \frac{7\pi}{4}) = (3\sqrt{2}, -4)$Plotting these points,we see they form a rectangle with vertices at $(3\sqrt{2}, 4), (-3\sqrt{2}, 4), (-3\sqrt{2}, -4),$ and $(3\sqrt{2}, -4)$.The sides are parallel to the coordinate axes,and the lengths are $6\sqrt{2}$ and $8$. Thus,it is a rectangle.

Let the points $P_1\left(\frac{\pi}{4}\right), P_2\left(\frac{3 \pi}{4}\right), P_3\left(\frac{5 \pi}{4}\right)$ and $P_4\left(\frac{7 \pi}{4}\right)$ given in parametric form,lie on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$. Then these four points in that order form

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