A) \[{{n}^{2}}({{y}^{2}}-4)\]
B) \[{{n}^{2}}(4-{{y}^{2}})\]
C) \[{{n}^{2}}({{y}^{2}}+4)\]
D) None of these
Correct Answer: C
Solution :
\[x=\sec \theta -\cos \theta \] \[\Rightarrow \] \[\frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta ,\] \[y={{\sec }^{n}}\theta -{{\cos }^{n}}\theta \] \[\Rightarrow \] \[\frac{dy}{d\theta }=n{{\sec }^{n-1}}\theta \sec \theta \tan \theta +n{{\cos }^{n-1}}\theta \sin \theta \] \[\therefore \] \[\frac{dy}{dx}=n\frac{({{\sec }^{n}}\theta \tan \theta +{{\cos }^{n-1}}\theta \sin \theta )}{(\sec \theta \tan \theta +\sin \theta )}\] \[\Rightarrow \] \[\frac{dy}{dx}=n\frac{({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )\tan \theta }{(\sec \theta +\cos \theta )\tan \theta }\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{n\,({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}{(\sec \theta +\cos \theta )}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}^{2}}=\frac{n\{\,{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4\}}{{{(\sec \theta -\cos \theta )}^{2}}+4}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}^{2}}=\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\] \[\Rightarrow \] \[({{x}^{2}}+4){{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4)\]
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