# Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

If $x=a\left(t+\frac{1}{t}\right)$ and $y=a\left(t-\frac{1}{t}\right)$, prove that $\frac{d y}{d x}=\frac{x}{y}$

Solution:

as $x=a\left(t+\frac{1}{t}\right)$

Differentiating it with respect to $t$,

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{ad}}{\mathrm{dt}}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)$

$=\mathrm{a}\left(1-\frac{1}{\mathrm{t}^{2}}\right)$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}\left(\frac{\mathrm{t}^{2}-1}{\mathrm{t}^{2}}\right) \cdots \cdots(1)$

And $\mathrm{y}=\mathrm{a}\left(\mathrm{t}-\frac{1}{t}\right)$

Differentiating it with respect to $t$,

$\frac{d y}{d t}=\frac{a d}{d t}\left(t-\frac{1}{t}\right)$

$=a\left(1+\frac{1}{t^{2}}\right)$

$\frac{d y}{d t}=a\left(\frac{t^{2}+1}{t^{2}}\right) \ldots \ldots(2)$

Dividing equation (2) by (1),

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=a\left(\frac{t^{2}+1}{t^{2}}\right) \times \frac{t^{2}}{a\left(t^{2}-1\right)}$

$\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1}$

$\frac{d y}{d x}=\frac{x}{y}\left[\right.$ since, $\left.\frac{x}{y}=a\left(\frac{t^{2}+1}{t^{2}}\right) \times \frac{t^{2}}{a\left(t^{2}-1\right)}=\left(\frac{t^{2}+1}{t^{2}-1}\right)\right]$

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