Let θ be the acute angle between the tangents to the ellipse 9x2+1y2=1 and the circle x2+y2=3 at their point of intersection in the first quadrant. Then tanθ is equal to
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Official solution
The point of intersection of the curves \[{{{x^2}} \over 9} + {{{y^2}} \over 1} = 1\] and x2+y2=3 in the first quadrant is \[\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)\]
Now slope of tangent to the ellipse \[{{{x^2}} \over 9} + {{{y^2}} \over 1} = 1\] at \[\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)\] is
\[{m_1} = - {1 \over {3\sqrt 3 }}\]
And slope of tangent to the circle at \[\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)\] is m2 \[ = - \sqrt 3 \]
So, if angle between both curves is \[\theta\] then
\[\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|\]
\[ = {2 \over {\sqrt 3 }}\]
Option (b)