If \(\mathrm{x}=\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)\) and \(\mathrm{y}=\mathrm{b}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)\), then \(\frac{\mathrm{dy}}{\mathrm{dx}}=\)

\(\mathrm{x}=\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right) \text { and } \mathrm{y}=\mathrm{b}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right) \)
\( \therefore \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right) \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right) \)
\( \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right)}{\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right)}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{t}^2-1}{\mathrm{t}^2+1}\right)\)
Now \(\mathrm{x}=\mathrm{a}\left(\frac{\mathrm{t}^2-1}{\mathrm{t}}\right)\) and \(\mathrm{y}=\mathrm{b}\left(\frac{\mathrm{t}^2+1}{\mathrm{t}}\right)\)
...[From (1)]
\(
\therefore\left(\mathrm{t}^2-1\right)=\left(\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{t} \text { and }\left(\mathrm{t}^2+1\right)=\left(\frac{\mathrm{y}}{\mathrm{b}}\right)(\mathrm{t})
\)
\(\therefore\) Eq. (2) becomes
\(
\frac{d y}{d x}=\left(\frac{b}{a}\right)\left(\frac{x}{a}\right)(t) \times \frac{1}{\left(\frac{y}{b}\right) \times t}=\frac{b^2 x}{a^2 y}
\)