Question

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

• A. $\frac{1}{2}\sqrt{\frac{5}{3}}$
• B. $\frac{1}{2}\sqrt{\frac{11}{3}}$
• C. $\frac{1}{3}\sqrt{\frac{11}{3}}$
• D. $\sqrt{\frac{5}{6}}$

Answer 1

The correct option is B

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

Explanation for correct option:

Let $m$ be the slope , the length of major axis $2a$, length of minor axis $2b$ and $e$ the eccentricity.

Minor axis $=\frac{4}{\sqrt{3}}$

$$\begin{gathered} \Rightarrow 2b=\frac{4}{\sqrt{3}} \\ \Rightarrow b=\frac{2}{\sqrt{3}} \end{gathered}$$

Equation of the tangent $\Rightarrow y=mx\pm \sqrt{a^2{m}^2+b^2}$

Comparing with $y=-\frac{x}{6}+\frac{4}{3}$

We get $m=-\frac{1}{6}$ and $\sqrt{a^2{m}^2+b^2}=\frac{4}{3}$

So, $a^2{m}^2+b^2=\frac{16}{9}$

$$\begin{gathered} \Rightarrow \frac{a^2}{36}+\frac{4}{3}=\frac{16}{9} \\ \Rightarrow \frac{a^2}{36}=\frac{16}{9}-\frac{4}{3} \\ \Rightarrow \frac{a^2}{36}=\frac{4}{9} \\ \Rightarrow a^2=16 \\ \therefore e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{3\times 16}}=\sqrt{\frac{11}{12}}=\frac{1}{2}\sqrt{\frac{11}{3}} \end{gathered}$$

Hence, the correct option is Option B.