Let the foci of the ellipse (x²)/16+(y²)/7=1 and the hyperbola (x²)/144-(y²)/(α)=1/25 coincide. Then the length of the latus rectum of the hyperbola is:

Coordinate Geometry: JEE Main 2022 Mathematics question

JEE Main 2022 · 25 Jul, Evening shift · Q3974

JEE Main 202225 Jul, Evening shiftMathematicsMediumCoordinate Geometry478× since 2002

Let the foci of the ellipse

\frac{x^{2}}{16}+\frac{y^{2}}{7}=1

and the hyperbola

\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}

coincide. Then the length of the latus rectum of the hyperbola is :

JEE Main 2022 · 25 Jul, Evening shift · Q3974

JEE Main 202225 Jul, Evening shiftMathematicsMediumCoordinate Geometry478× since 2002

Let the foci of the ellipse

\frac{x^{2}}{16}+\frac{y^{2}}{7}=1

and the hyperbola

\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}

coincide. Then the length of the latus rectum of the hyperbola is :