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 $=\pi$