Let 0 < θ < (π)/2. If the eccentricity of the hyperbola (x²)/(cos ²θ) - (y²)/(sin ²θ) = 1 is greater than 2, then the length of its latus rectum lies in the interval:

Coordinate Geometry: JEE Main 2019 Mathematics question

JEE Main 2019 · 9 Jan, Morning shift · Q3962

JEE Main 20199 Jan, Morning shiftMathematicsHardCoordinate Geometry478× since 2002

Let

0 < \theta < {\pi \over 2}

. If the eccentricity of the hyperbola

{{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}

= 1 is greater than 2, then the length of its latus rectum lies in the interval :

JEE Main 2019 · 9 Jan, Morning shift · Q3962

JEE Main 20199 Jan, Morning shiftMathematicsHardCoordinate Geometry478× since 2002

Let

0 < \theta < {\pi \over 2}

. If the eccentricity of the hyperbola

{{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}

= 1 is greater than 2, then the length of its latus rectum lies in the interval :