Question:
If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.
$x=a \sec \theta, y=b \tan \theta$
Solution:
The given equations are $x=a \sec \theta$ and $y=b \tan \theta$
Then, $\frac{d x}{d \theta}=a \cdot \frac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta$
$\frac{d y}{d \theta}=b \cdot \frac{d}{d \theta}(\tan \theta)=b \sec ^{2} \theta$
$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}=\frac{b}{a} \sec \theta \cot \theta=\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} \operatorname{cosec} \theta$