If the maximum distance of normal to the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 2$, from the origin is 1, then the eccentricity of the ellipse is :

Equation of normal is <br/><br/>$2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^{2}$ <br/><br/>Distance from $(0,0)=\frac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta}}$ <br/><br/>Distance is maximum if <br/><br/>$4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta$ is minimum <br/><br/>$\Rightarrow \tan ^{2} \theta=\frac{\mathrm{b}}{2}$ <br/><br/>$\Rightarrow \frac{4-b^{2}}{\sqrt{4 \cdot \frac{b+2}{2}+b^{2} \cdot \frac{b+2}{b}}}=1$ <br/><br/>$\Rightarrow 4-b^{2}=(b+2) \Rightarrow b^{2}+b-2=0$ <br/><br/>$\Rightarrow(b+2)(b-1)=0$ <br/><br/>$\Rightarrow b=1$ <br/><br/>$\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$