A) \[\frac{-y}{x}\]
B) \[\frac{y}{x}\]
C) \[\frac{-x}{y}\]
D) \[\frac{x}{y}\]
Correct Answer: C
Solution :
\[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\] and \[y=\frac{2t}{1+{{t}^{2}}}\] Put \[t=\tan \theta \] in both the equations, we get \[x=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \] and \[y=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta \]. Differentiating both the equations, we get \[\frac{dx}{d\theta }=-2\sin 2\theta \] and \[\frac{dy}{d\theta }=2\cos 2\theta .\] Therefore \[\frac{dy}{dx}=-\frac{\cos 2\theta }{\sin 2\theta }=-\frac{x}{y}\].
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