If the foci of a hyperbola are same as that of the ellipse (x²)/9+(y²)/25=1 and the eccentricity of the hyperbola is 15/8 times the eccentricity of the ellipse, then the smaller focal distance of the point (√(2), 14/3 √(2/5)) on the hyperbola, is equal to

Coordinate Geometry: JEE Main 2024 Mathematics question

JEE Main 2024 · 31 Jan, Morning shift · Q3979

JEE Main 202431 Jan, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002

If the foci of a hyperbola are same as that of the ellipse

\frac{x^2}{9}+\frac{y^2}{25}=1

and the eccentricity of the hyperbola is

\frac{15}{8}

times the eccentricity of the ellipse, then the smaller focal distance of the point

\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)

on the hyperbola, is equal to