A) \[-y/x\]
B) \[y/x\]
C) \[-x/y\]
D) \[x/y\]
Correct Answer: C
Solution :
Equation \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2t}{1+{{t}^{2}}}\] Put \[t=\tan \theta \Rightarrow x=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[\Rightarrow \]\[x=\cos 2\theta \] \[\Rightarrow \]\[y=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }\Rightarrow y=\sin 2\theta ,x=\cos 2\theta \] \[\frac{dx}{d\theta }=-2\sin 2\theta ,\frac{dy}{d\theta }=2\cos 2\theta \] \[\frac{dy}{dx}=-\frac{2\cos 2\theta }{2\sin 2\theta }=-x/y\]
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