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(B) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $ax \sec 	heta - b

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$\frac{\sqrt{3}}{2}$

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(B) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is $ax \sec 	heta - by \operatorname{cosec} 	heta = a^2 - b^2$.Here $a^2 = 4$,so the equation is $2x \sec 	heta - by \operatorname{cosec} 	heta = 4 - b^2$.The perpendicular distance $d$ from the origin $(0,0)$ is $d = \frac{|4 - b^2|}{\sqrt{4 \sec^2 	heta + b^2 \operatorname{cosec}^2 	heta}}$.To maximize $d$,we minimize $f(	heta) = 4 \sec^2 	heta + b^2 \operatorname{cosec}^2 	heta = 4(1 + 	an^2 	heta) + b^2(1 + \cot^2 	heta) = (4 + b^2) + 4 	an^2 	heta + b^2 \cot^2 	heta$.Using $AM$-$GM$ inequality,$4 	an^2 	heta + b^2 \cot^2 	heta \geq 2 \sqrt{4 	an^2 	heta \cdot b^2 \cot^2 	heta} = 4b$.The minimum value is $(4 + b^2 + 4b) = (2 + b)^2$.Thus,$d_{max} = \frac{4 - b^2}{\sqrt{(2 + b)^2}} = \frac{(2 - b)(2 + b)}{2 + b} = 2 - b$.Given $d_{max} = 1$,we have $2 - b = 1$,so $b = 1$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

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

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