The length of the minor axis (along y-axis) of an ellipse in
Question:

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:

  1. (1) $\frac{1}{2} \sqrt{\frac{11}{3}}$

  2. (2) $\sqrt{\frac{5}{6}}$

  3. (3) $\frac{1}{2} \sqrt{\frac{5}{3}}$

  4. (4) $\frac{1}{3} \sqrt{\frac{11}{3}}$


Correct Option: 1

Solution:

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 $(e)=\sqrt{1-\frac{b^{2}}{a^{2}}}$

$\Rightarrow \quad e=\sqrt{1-\frac{4}{3 \times 16}}=\sqrt{\frac{11}{12}}=\frac{1}{2} \sqrt{\frac{11}{3}}$

 

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