Question

If for the ellipse $\left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}\right)+\left(\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}\right)=1$, the y-axis is the minor axis and the length of the latus rectum is one half of the length of its minor axis, then its eccentricity is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{3}{4}$
• E. $\frac{3}{5}$

Answer 1

The correct option is C

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

Explanation for the correct answer:

Given, the equation of the ellipse is,

$\left(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}\right)+\left(\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}\right)=1$

given, the length of latus rectum = one half of the length of its minor axis.

Length of the latus rectum$=\frac{2{\mathrm{b}}^2}{\mathrm{a}}$

Length of minor axis$=2\mathrm{b}$

According to the problem,

$\begin{array}{rcl}\frac{2{\mathrm{b}}^2}{\mathrm{a}}& =& \frac{1}{2}\left(2\mathrm{b}\right)\\ \frac{2{\mathrm{b}}^2}{\mathrm{a}}& =& \mathrm{b}\\ \mathrm{b}& =& \frac{\mathrm{a}}{2}\\ \mathrm{a}& =& 2\mathrm{b}\end{array}$

Now, the eccentricity of ellipse is,

$\begin{array}{rcl}\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}& =& \sqrt{1-\frac{{\mathrm{b}}^2}{{\left(2\mathrm{b}\right)}^2}}\\ & =& \sqrt{1-\frac{{\mathrm{b}}^2}{4{\mathrm{b}}^2}}\\ & =& \sqrt{\frac{4{\mathrm{b}}^2-{\mathrm{b}}^2}{4{\mathrm{b}}^2}}\\ & =& \sqrt{\frac{3{\mathrm{b}}^2}{4{\mathrm{b}}^2}}\\ & =& \frac{\sqrt{3}}{2}\end{array}$

Hence, option C is correct.