Question

If $x$ and $y$ are connected parametrically by the given equation, then without eliminating the parameter, find $\frac{dy}{dx}$.
$x=a\sec \theta ,y=b\tan \theta$

Answer 1

$x=a\sec \theta ,y=b\tan \theta$
$\Rightarrow \frac{dx}{d\theta }=a.\frac{d}{d\theta }\left(\sec \theta \right)=a\sec \theta \tan \theta$
& $\frac{dy}{d\theta }=b.\frac{d}{d\theta }\left(\tan \theta \right)=b{\sec }^2\theta$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{b{\sec }^2\theta }{a\sec \theta \tan \theta }=\frac{b}{a}\sec \theta \cot \theta =\frac{b\cos \theta }{a\cos \theta \sin \theta }=\frac{b}{a}\times \frac{1}{\sin \theta }=\frac{b}{a}co\sec \theta$