Let the eccentricity of the ellipse x² + a²y² = 25a² be b times the eccentricity of the hyperbola x² - a²y² = 5, where a is the minimum distance between the curves y = eˣ and y = log e x. Then a² + 1/(b²) is equal to:

Coordinate Geometry: JEE Main 2022 Mathematics question

JEE Main 2022 · 30 Jun, Morning shift · Q3762

JEE Main 202230 Jun, Morning shiftMathematicsHardCoordinate Geometry478× since 2002

Let the eccentricity of the ellipse

{x^2} + {a^2}{y^2} = 25{a^2}

be b times the eccentricity of the hyperbola

{x^2} - {a^2}{y^2} = 5

, where a is the minimum distance between the curves y = e^x and y = log_ex. Then

{a^2} + {1 \over {{b^2}}}

is equal to :