A) \[\sqrt{2}\]
B) \[\frac{2}{\sqrt{3}}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[\frac{\sqrt{3}}{4}\]
Correct Answer: B
Solution :
\[{{x}^{2}}+3{{y}^{2}}=9\] \[\Rightarrow 2x+6y\frac{dy}{dx}=0........\]Differentiating w.r.t \[\Rightarrow \frac{dy}{dx}=\frac{-x}{3y}\] Equation of normal is \[-\frac{dx}{dy}=\frac{3y}{x}\] \[{{\left. \frac{dx}{dy} \right|}_{(3\cos \theta ,\sqrt{3}\sin \theta )}}=\frac{3\sqrt{3}\sin \theta }{-3\cos \theta }=\sqrt{3}\tan \theta ={{m}_{1}}\] \[{{\left. \frac{dx}{dy} \right|}_{(-3\sin \theta ,\sqrt{3}\cos \theta )}}=\frac{3\sqrt{3}\cos \theta }{-3sin\theta }=-\sqrt{3}\cot \theta ={{m}_{2}}\] \[\beta \] is the angle between the normal to the ellipse \[(i)\], then \[\tan \beta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[=\left| \frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-3\tan \theta \cot \theta } \right|\] \[=\left| \frac{\sqrt{3}\tan \theta +\sqrt{3}\cot \theta }{1-3} \right|\] \[\tan \beta =\frac{\sqrt{3}}{2}|\tan \theta +\cot \theta |\] \[\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}|\tan \theta +\cot \theta |\] \[\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}\left| \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \right|\] \[\frac{1}{\cot \beta }=\frac{\sqrt{3}}{2}\left| \frac{1}{\sin \theta cos\theta } \right|\] \[\frac{1}{\cot \beta }=\frac{\sqrt{3}}{\sin 2\theta }\] \[\Rightarrow \frac{2\cot \beta }{\sin 2\theta }=\frac{2}{\sqrt{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
