If $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2t}{1 + t^{2}},then\ \frac{dy}{dx}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$\because x = \frac{1 - t^{2}}{1 + t^{2}}\ and\ y = \frac{2t}{1 + t^{2}}$
Put $t = 	an	heta$ in both the equations, we get
$x = \frac{1 - 	an^{2}	heta}{1 + 	an^{2}	heta} = \cos{2	heta}\ldots(i)$
and $y = \frac{2	an	heta}{1 + 	an^{2}	heta} = \sin{2	heta\ldots(ii)}$
On differentiating both the Eqs.(i) and (ii), we get
$\frac{dx}{d	heta} = - 2\sin{2	heta\ and\ \frac{dy}{d	heta} = 2\cos{2	heta}}$
Therefore, $\frac{dy}{dx} = \frac{\frac{dy}{d	heta}}{\frac{dx}{d	heta}} = - \frac{\cos{2	heta}}{\sin{2	heta}} = - \frac{x}{y}$

Q. If $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2t}{1 + t^{2}},then\ \frac{dy}{dx}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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