JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Ellipse

The normal to the ellipse at the point with parameter $\theta$ is $2x\sec \theta -by\csc \theta =4-b^2$, so its distance from the origin is $\frac{4-b^2}{\sqrt{4{\sec }^2\theta +b^2{\csc }^2\theta }}$. The denominator $\frac{4}{{\cos }^2\theta }+\frac{b^2}{{\sin }^2\theta }$ has least value $(2+b{)}^2$, so the greatest distance is $\frac{4-b^2}{2+b}=2-b$. Setting $2-b=1$ gives $b=1$. With $a=2$ and $b=1$, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$.