JEE Main 202331 Jan, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002If the maximum distance of normal to the ellipse x24+y2b2=1,b<2\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 24x2+b2y2=1,b<2, from the origin is 1, then the eccentricity of the ellipse is :AOption A. 34\frac{\sqrt{3}}{4}43BOption B. 12\frac{1}{2}21COption C. 12\frac{1}{\sqrt{2}}21DOption D. 32\frac{\sqrt{3}}{2}23CheckSkip · show solution
