Let the foci and length of the latus rectum of an ellipse (x²)/(a²)+(y²)/(b²)=1, a>b b e(± 5,0) and √(50), respectively. Then, the square of the eccentricity of the hyperbola (x²)/(b²)-(y²)/(a² b²)=1 equals

Coordinate Geometry: JEE Main 2024 Mathematics question

JEE Main 2024 · 31 Jan, Morning shift · Q42

JEE Main 202431 Jan, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002

Let the foci and length of the latus rectum of an ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)

and

\sqrt{50}

, respectively. Then, the square of the eccentricity of the hyperbola

\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1

equals

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