JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Ellipse

For the ellipse $\frac{x^2}{b^2}+\frac{y^2}{4a^2}=1$, the equation of a tangent at parameter $\theta$ is $\frac{x\cos \theta }{b}+\frac{y\sin \theta }{2a}=1$. This tangent meets the x-axis at $A(b\sec \theta ,0)$ and the y-axis at $B(0,2a\csc \theta )$. The area of triangle OAB is $\Delta =\frac{1}{2}|b\sec \theta ||2a\csc \theta |=ab|\sec \theta \csc \theta |$. To minimize, we use AM-GM: ${\sec }^2\theta +{\csc }^2\theta \ge 2\sec \theta \csc \theta$, with equality when $\sec \theta =\csc \theta$, i.e., $\theta =\pi /4$. At this angle, ${\Delta }_{\min }=ab\cdot \sqrt{2}\cdot \sqrt{2}=2ab$. Thus $k=2$.