Question

If $x=a(t-\frac{1}{t}),y=a(t+\frac{1}{t})$, then show that $\frac{dy}{dx}=\frac{x}{y}$

Answer 1

Let $x=a(t-\frac{1}{t})$ ... (i)
$\Rightarrow \frac{dx}{dt}=a(1+\frac{1}{t^2})$
and $y=a(t+\frac{1}{t})$ ... (ii)
$\Rightarrow \frac{dy}{dt}=a(1-\frac{1}{t^2})$
Therefore, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$=\frac{a(1-\frac{1}{t^2})}{a(1+\frac{1}{t^2})}$
$=\frac{a(t^2-1)}{a(t^2+1)}$
$=\frac{a(t-\frac{1}{t})}{a(t+\frac{1}{t})}$
$=\frac{x}{y}$ ....[From (i) and (ii)]