JEE Main 202131 Aug, Evening shiftMathematicsMediumCoordinate Geometry478× since 2002An angle of intersection of the curves, x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1a2x2+b2y2=1 and x² + y² = ab, a > b, is :AOption A. tan−1(a+bab){\tan ^{ - 1}}\left( {{{a + b} \over {\sqrt {ab} }}} \right)tan−1(aba+b)BOption B. tan−1(a−b2ab){\tan ^{ - 1}}\left( {{{a - b} \over {2\sqrt {ab} }}} \right)tan−1(2aba−b)COption C. tan−1(a−bab){\tan ^{ - 1}}\left( {{{a - b} \over {\sqrt {ab} }}} \right)tan−1(aba−b)DOption D. tan−1(2ab){\tan ^{ - 1}}\left( {2\sqrt {ab} } \right)tan−1(2ab)CheckSkip · show solution
