If the maximum distance of normal to the ellipse x ^ 2 4 + y ^ 2 b...

Given:

$\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

Equation of normal to the ellipse is

$2xsec\theta - bycosec\theta = 4 - {b}^{2}$

Distance of normal from origin is

$d = \left|\dfrac{4 - {b}^{2}}{\sqrt{4\sec^{2}\theta + {b}^{2}{cosec}^{2}\theta}}\right|$

For maximum distance, denominator must be minimum.

$\sqrt{4\sec^{2}\theta + {b}^{2}{cosec}^{2}\theta}$

$= \sqrt{4 + 4\tan^{2}\theta + {b}^{2} + {b}^{2}\cot^{2}\theta}$

Now,

$\dfrac{4\tan^{2}\theta + {b}^{2}\cot^{2}\theta}{2} \ge \sqrt{4\tan^{2}\theta \times {b}^{2}\cot^{2}\theta}$

$⇒ 4\tan^{2}\theta + {b}^{2}\cot^{2}\theta \ge 4b$

$⇒ 4 + {b}^{2} + 4\tan^{2}\theta + {b}^{2}\cot^{2}\theta \ge 4 + {b}^{2} + 4b$

$⇒ 4 + {b}^{2} + 4\tan^{2}\theta + {b}^{2}\cot^{2}\theta \ge {\left(2 + b\right)}^{2}$

So, minimum value of $\sqrt{4\sec^{2}\theta + {b}^{2}{cosec}^{2}\theta}$ is $\sqrt{{\left(2 + b\right)}^{2}} = b + 2$

So,

$d = \left|\dfrac{4 - {b}^{2}}{2 + b}\right| = 1$

$⇒ \left|2 - b\right| = 1$

$⇒ 2 - b = \pm 1$

$⇒ b = 1, 3$

But $b < 2$, so $b = 1$, hence

$e = \sqrt{1 - \dfrac{1}{4}} = \dfrac{\sqrt{3}}{2}$