JEE Main 202227 Jun, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002Let the eccentricity of an ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1a2x2+b2y2=1, a>ba > ba>b, be 14{1 \over 4}41. If this ellipse passes through the point (−425,3)\left( { - 4\sqrt {{2 \over 5}} ,3} \right)(−452,3), then a2+b2{a^2} + {b^2}a2+b2 is equal to :AOption A. 29BOption B. 31COption C. 32DOption D. 34CheckSkip · show solution
