If x = a ( t - 1 t ) , y = b ( t + 1 t ) , then d y d x =

Question

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{b^{2} x}{a^{2} y}$

Explanation:

We have

$x = a (t - \frac{1}{t})$

$y = b (t + \frac{1}{t})$

On differentiating both equations with respect to $t$ , we get

$\frac{d x}{d t} = a [\frac{d t}{d t} - \frac{d (t^{- 1})}{d t}]$

$\frac{d y}{d t} = b [\frac{d t}{d t} + \frac{d (t^{- 1})}{d t}]$

$\frac{d x}{d t} = a [1 + \frac{1}{t^{2}}]$ $\dots (i)$

$\frac{d y}{d t} = b [1 - \frac{1}{t^{2}}]$ $\dots (i i)$

On dividing equation (ii) by equation (i), we get

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b [\frac{t^{2} - 1}{t^{2}}]}{a [\frac{t^{2} + 1}{t^{2}}]}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (t^{2} - 1)}{a (t^{2} + 1)} = \frac{b (t - \frac{1}{t})}{a (t + \frac{1}{t})}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (\frac{x}{a})}{a (\frac{y}{b})} = \frac{b^{2} x}{a^{2} y}$

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