$If x = \frac{2 at}{1 + t^{2}}, y = \frac{b (1 - t^{2})}{1 + t^{2}},$ then $\frac{dy}{dx}$

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$\frac{y}{x}$

$- \frac{b^{2} x}{a^{2} y}$

$- \frac{x}{y}$

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

answer is C.

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Detailed Solution

Given $x = \frac{2 at}{1 + t^{2}}, y = \frac{b (1 - t^{2})}{1 + t^{2}},$

Differentiate both $x$ and $y$ with respect to $t$

$\begin{array}{rcl} \frac{d x}{d t} & = & \frac{d}{d t} (\frac{2 a t}{1 + t^{2}}) \\ = & 2 a \frac{(1 + t^{2}) 1 - t (2 t)}{{(1 + t^{2})}^{2}} \\ = & 2 a \frac{1 + t^{2} - 2 t^{2}}{{(1 + t^{2})}^{2}} \\ = & 2 a \frac{1 - t^{2}}{{(1 + t^{2})}^{2}} \end{array}$

$\begin{array}{rcl} \frac{d y}{d t} & = & \frac{d}{d t} (\frac{b (1 - t^{2})}{1 + t^{2}}) \\ = & b \frac{(1 + t^{2}) (- 2 t) - (1 - t^{2}) 2 t}{{(1 + t^{2})}^{2}} \\ = & \frac{- 2 b t (1 + t^{2} + 1 - t^{2})}{{(1 + t^{2})}^{2}} \\ = & \frac{- 4 b t}{{(1 + t^{2})}^{2}} \end{array}$

Use the concept of parametric differentiation

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d t} \cdot \frac{d t}{d x} \\ = & - \frac{4 b t}{{(1 + t^{2})}^{2}} \cdot \frac{{(1 + t^{2})}^{2}}{2 a (1 - t^{2})} \\ = & \frac{- 2 b t}{a (1 - t^{2})} \\ = & \frac{- b \frac{x}{a}}{a \frac{y}{b}} \\ = & \frac{- b^{2} x}{a^{2} y} \end{array}$