Question

If the points $(\sqrt{3}\text{sin}\theta ,\sqrt{4}\text{cos}\theta )$ where $\theta \in R$, lies outside the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$, then the value of $\theta$ 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 $\theta \in R$

$S<0$ is the condition for the point to lie outside the hyperbola.
$\Rightarrow \frac{3{\text{sin}}^2\theta }{4}-\frac{4{\text{cos}}^2\theta }{5}-1<0$
$\Rightarrow 15{\text{sin}}^2\theta -16{\text{cos}}^2\theta -20<0$
$\Rightarrow 15(1-{\text{cos}}^2\theta )-16{\text{cos}}^2\theta -20<0$
$\Rightarrow 15-15{\text{cos}}^2\theta -16{\text{cos}}^2\theta -20<0$
$\Rightarrow -5-31{\text{cos}}^2\theta <0$
$\Rightarrow 31{\text{cos}}^2\theta +5>0$
$\therefore \theta \in R(\because 0\le {\text{cos}}^2\theta \le 1)$