Question

If the eccentricity of the hyperbola $\frac{x^{2}}{\left(1 + sin \theta \right)^{2}}-\frac{y^{2}}{\left(cos\right)^{2} ⁡ \theta }=1$ is $\frac{2}{\sqrt{3}},$ then the sum of all the possible values of $\theta $ is (where, $\theta \in \left(0 , \pi \right)$ )

• A. $\frac{5 \pi }{4}$
• B. $\frac{2 \pi }{3}$
• C. $\frac{7 \pi }{4}$
• D. $\pi $

Answer 1

Answer: (D)

$e^{2}=1+\frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}=\frac{4}{3}$
$\Rightarrow \frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}=\frac{1}{3}$
$\Rightarrow \left[\frac{\cos \theta}{1+\sin \theta}\right]^{2}=\frac{1}{3}$
$\Rightarrow \left[\frac{\sin \left(\frac{\pi}{2}-\theta\right)}{1+\cos \left(\frac{\pi}{2}-\theta\right)}\right]^{2}=\frac{1}{3}$
$\Rightarrow \tan ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\frac{1}{3}$
$\frac{\pi}{4}-\frac{\theta}{2}=\frac{-\pi}{6}, \frac{\pi}{6}$
$\Rightarrow \frac{\theta}{2}=\frac{\pi}{4}-\frac{\pi}{6}, \frac{\pi}{4}+\frac{\pi}{6}$
$\theta=\frac{\pi}{2} \pm \frac{\pi}{3}=\frac{\pi}{6}, \frac{5 \pi}{6}$
Hence, the required sum $=\pi$