Let $\theta$ be the acute angle between the tangents to the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ and the circle ${x^2} + {y^2} = 3$ at their point of intersection in the first quadrant. Then tan$\theta$ is equal to :

The point of intersection of the curves ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ and ${x^2} + {y^2} = 3$ in the first quadrant is $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$<br><br>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 <br><br>${m_1} = - {1 \over {3\sqrt 3 }}$<br><br>And slope of tangent to the circle at $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$ is m₂ $= - \sqrt 3$<br><br>So, if angle between both curves is $\theta$ then<br><br>$$\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|$$<br><br>$= {2 \over {\sqrt 3 }}$<br><br>Option (b)