Question:
$x=t+\frac{1}{t}, y=t-\frac{1}{t}$
Solution:
Given,
x = t + 1/t, y = t – 1/t
Differentiating both the parametric functions w.r.t θ
$\frac{d x}{d t}=1-\frac{1}{t^{2}}, \frac{d y}{d t}=1+\frac{1}{t^{2}}$
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1+\frac{1}{t^{2}}}{1-\frac{1}{t^{2}}}=\frac{t^{2}+1}{t^{2}-1}$
Thus, $\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1}$.