If $x=a\left(t+\frac{1}{t}\right)$ and $y=a\left(t-\frac{1}{t}\right)$, prove that $\frac{d y}{d x}=\frac{x}{y}$
We have, $x=a\left(t+\frac{1}{t}\right)$ and $y=a\left(t-\frac{1}{t}\right)$
$\Rightarrow \frac{d x}{d t}=a \frac{d}{d t}\left(t+\frac{1}{t}\right)$ and $\frac{d y}{d t}=a \frac{d}{d t}\left(t-\frac{1}{t}\right)$
$\Rightarrow \frac{d x}{d t}=a\left(1-\frac{1}{t^{2}}\right)$ and $\frac{d y}{d t}=a\left(1+\frac{1}{t^{2}}\right)$
$\Rightarrow \frac{d x}{d t}=a\left(\frac{t^{2}-1}{t^{2}}\right)$ and $\frac{d y}{d t}=a\left(\frac{t^{2}+1}{t^{2}}\right)$
$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a\left(t^{2}+1\right)}{t^{2}} \times \frac{t^{2}}{a\left(t^{2}-1\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{a\left(t^{2}+1\right)}{t} \times \frac{t}{a\left(t^{2}-1\right)}$
$\Rightarrow \frac{d y}{d x}=a\left(t+\frac{1}{t}\right) \times \frac{1}{a\left(t-\frac{1}{t}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{x}{y}$
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