If \beta is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos heta , \sqrt 3 \sin heta ) and (- 3 sin heta , \sqrt 3 \,\cos heta ); heta \in \left( {0,{\pi \over 2}} ight); then {{2\,\cot \beta } \over {\sin 2 heta }} is equal to :

Question

If \beta is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos 	heta , \sqrt 3 \sin 	heta ) and (- 3 sin 	heta , \sqrt 3 \,\cos 	heta ); 	heta \in \left( {0,{\pi \over 2}} ight); then {{2\,\cot \beta } \over {\sin 2	heta }} is equal to :

Accepted Answer

Answer: A. {2 \over {\sqrt 3 }} Since, x2 + 3y2 = 9 \Rightarrow 2x + 6y {{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{ - x} \over {3y}} Slope of normal is - {{dx} \over {dy}} = {{3y} \over x} \Rightarrow {\left( { - {{dx} \over {dy}}} ight)_{\left( {3\cos 	heta ,\sqrt 3 \sin 	heta } ight)}} = {{3\sqrt 3 \sin 	heta } \over {3\cos 	heta }} = \sqrt 3 	an 	heta = m1 & {\left( { - {{dx} \over {dy}}} ight)_{\left( { - 3\sin 	heta ,\sqrt 3 \cos 	heta } ight)}} = {{3\sqrt 3 \cos 	heta } \over { - 3\sin 	heta }} = - \sqrt 3 \cot 	heta = m2 As, \beta is the angle between the normals to the given ellipse then tan\beta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} ight| = \left| {{{\sqrt 3 	an 	heta + \sqrt 3 \cot 	heta } \over {1 - 3	an 	heta \cot 	heta }}} ight| = \left| {{{\sqrt 3 	an 	heta + \sqrt 3 \cot 	heta } \over {1 - 3}}} ight| So, tan \beta = {{\sqrt 3 } \over 2} \left| {	an 	heta + \cot 	heta } ight| \Rightarrow {1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin 	heta } \over {\cos 	heta }} + {{\cos 	heta } \over {\sin 	heta }}} ight| \Rightarrow {1 \over {\cot \beta }} = {{\sqrt 3 } \over 2} \left| {{1 \over {\sin 	heta \cos 	heta }}} ight| \Rightarrow {1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2	heta }} \Rightarrow {{2\cot \beta } \over {\sin 2	heta }} = {2 \over {\sqrt 3 }}

Answer

Explanation

More JEE Math questions