Let the points $\mathrm{P}_1\left(\frac{\pi}{4}\right), \mathrm{P}_2\left(\frac{3 \pi}{4}\right), \mathrm{P}_3\left(\frac{5 \pi}{4}\right)$ and $\mathrm{P}_4\left(\frac{7 \pi}{4}\right)$ 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

• Correct answer
  A

  a rectangle
• B

  a square
• C

  a parallelogram
• D

  a rhombus

Solution

Since, $\frac{x^2}{9}-\frac{y^2}{16}=1$ ...(i) $\Rightarrow \quad a=3, b=4$ Parametric equation for hyperbola (i) is . $x=a \sec \theta, y=b \tan \theta$ $\Rightarrow \quad x=3 \sec \theta, y=4 \tan \theta$ $\begin{aligned} & P_1\left(\frac{\pi}{4}\right)=\left(3 \sec \frac{\pi}{4}, 4 \tan \frac{\pi}{4}\right)=(3 \sqrt{2}, 4) \\ & P_2\left(\frac{3 \pi}{4}\right)=\left(3 \sec \frac{3 \pi}{4}, 4 \tan \frac{3 \pi}{4}\right)=(-3 \sqrt{2},-4) \\ & P_3\left(\frac{5 \pi}{4}\right)=\left(3 \sec \frac{5 \pi}{4}, 4 \tan \frac{5 \pi}{4}\right)=(-3 \sqrt{2}, 4) \\ & P_4\left(\frac{7 \pi}{4}\right)=\left(3 \sec \frac{7 \pi}{4}, 4 \tan \frac{7 \pi}{4}\right)=(3 \sqrt{2},-4)\end{aligned}$
$P_1 P_3=P_2 P_4=\sqrt{72+0}=\sqrt{72}$ $\begin{aligned} & P_2 P_3=P_4 P_1=\sqrt{0+16}=4 \\ & P_1 P_2=P_3 P_4=\sqrt{18+64}=\sqrt{82}\end{aligned}$ $\therefore \mathrm{P}_1 \mathrm{P}_2 \mathrm{P}_3 \mathrm{P}_4$ is a rectangle.

Hence, the correct answer is option: 'A'.