Question

If the tangent at the point $(2\sec \theta ,3\tan \theta )$ of the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1,$ is parallel to $3x-y+4=0$, then find the value of $\theta$

Answer 1

Equation of tangent at $(2\sec \theta ,3\tan \theta )$ is

$\frac{x}{2}\sec \theta -\frac{y}{3}\tan \theta =1$ .......$\left(1\right)$

$\Rightarrow \frac{3}{2}\left(\frac{\sec \theta }{\tan \theta }\right)=3$

$\Rightarrow \frac{\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }}=2$

$\Rightarrow \frac{1}{\sin \theta }=2$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\therefore \theta ={30}^{\circ }$