Question

The range of $\theta$ for which the point $(\sqrt{3}\sin \theta ,\sqrt{4}\cos \theta )$ lies outside $\frac{x^2}{4}-\frac{y^2}{5}=1$ is

• A. $\theta \in R$
• B. $\theta \in R-\left\{\frac{\pi }{4}\right\}$
• C. $\theta \in R-\{\frac{\pi }{2},0\}$
• D. $\theta \in R-\{\frac{\pi }{4},0,\frac{\pi }{2}\}$

Answer 1

The correct option is A $\theta \in R$

$S_1<0$ is the condition for the point to lie outside the hyperbola.
$\Rightarrow \frac{3{\sin }^2\theta }{4}-\frac{4{\cos }^2\theta }{5}-1<0\Rightarrow 15{\sin }^2\theta -16{\cos }^2\theta -20<0\Rightarrow 31{\sin }^2\theta -36<0\Rightarrow {\sin }^2\theta <\frac{36}{31}\therefore \theta \in R(\because {\sin }^2\theta \in [0,1\left]\right)$