Let heta 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 heta is equal to :

Question

Let 	heta 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	heta is equal to :

Accepted Answer

Answer: B. {2 \over {\sqrt 3 }} 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}} ight) 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}} ight) is {m_1} = - {1 \over {3\sqrt 3 }} And slope of tangent to the circle at \left( {{3 \over 2},{{\sqrt 3 } \over 2}} ight) is m2 = - \sqrt 3 So, if angle between both curves is 	heta then 	an 	heta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} ight| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } ight)} ight)}}} ight| = {2 \over {\sqrt 3 }} Option (b)

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 :

Answer

Explanation

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