Question

If $P(\sqrt{2} \sec \theta, \sqrt{2} \tan \theta)$ is a point on the hyperbola whose distance from the origin is $\sqrt{6}$ where $P$ is in the first quadrant then $\theta=$

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}^2 \sec ^2 \theta+\sqrt{2}^2 \tan ^2 \theta=6 \Rightarrow 1+2 \tan ^2 \theta=3$
$\therefore \theta=\pi / 4$ for first quadrant