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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.Given the length of the minor axis is $2b = \frac{4}{\sqrt

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$\frac{1}{2} \sqrt{\frac{11}{3}}$

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.Given the length of the minor axis is $2b = \frac{4}{\sqrt{3}}$,so $b = \frac{2}{\sqrt{3}}$ and $b^2 = \frac{4}{3}$.The line $x + 6y = 8$ can be written as $y = -\frac{1}{6}x + \frac{4}{3}$.The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.Here $m = -\frac{1}{6}$ and $c = \frac{4}{3}$.Substituting the values: $(\frac{4}{3})^2 = a^2(-\frac{1}{6})^2 + \frac{4}{3}$.$\frac{16}{9} = \frac{a^2}{36} + \frac{4}{3}$.$\frac{a^2}{36} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9}$.$a^2 = \frac{4 	imes 36}{9} = 16$,so $a = 4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}} = \sqrt{\frac{11}{4 	imes 3}} = \frac{1}{2} \sqrt{\frac{11}{3}}$.

The length of the minor axis (along $y$-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}$. If this ellipse touches the line $x+6y=8$,then its eccentricity is

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