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,
$x+6 y=8$; then its eccentricity is:
Correct Option: 1
Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 ; a>b$
$2 b=\frac{4}{\sqrt{3}} \Rightarrow b=\frac{2}{\sqrt{3}}$
Equation of tangent $\equiv y=m x \pm \sqrt{a^{2} m^{2}+b^{2}}$
Comparing with $\equiv y=\frac{-x}{6}+\frac{4}{3}$
$m=\frac{-1}{6}$ and $a^{2} m^{2}+b^{2}=\frac{16}{9}$
$\Rightarrow \frac{a^{2}}{36}+\frac{4}{3}=\frac{16}{9} \Rightarrow \frac{a^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}$
$\Rightarrow a^{2}=16 \Rightarrow a=\pm 4$
Now, eccentricity of ellipse $(e)=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$\Rightarrow \quad e=\sqrt{1-\frac{4}{3 \times 16}}=\sqrt{\frac{11}{12}}=\frac{1}{2} \sqrt{\frac{11}{3}}$