Find the eccentricity of an ellipse whose latus rectum is one half of its major axis.
Let the equation of the required ellipse is
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ …(i)
It is given that,
Length of Latus Rectum $=\frac{1}{2}$ major Axis
We know that,
Length of Latus Rectum $=\frac{2 b^{2}}{a}$
and Length of Minor Axis $=2 \mathrm{a}$
So, according to the given condition,
$\frac{2 b^{2}}{a}=\frac{1}{2} \times 2 a$
$\Rightarrow \frac{2 b^{2}}{a}=a$
$\Rightarrow 2 b^{2}=a^{2} \ldots(i i)$
$\Rightarrow \mathrm{a}=\sqrt{2 \mathrm{~b}^{2}}$
$\Rightarrow \mathrm{a}=\mathrm{b} \sqrt{2}$
Now, we have to find the eccentricity
We know that,
Eccentricity, e $=\frac{c}{a}$ …(iii)
where, $c^{2}=a^{2}-b^{2}$
So, $c^{2}=2 b^{2}-b^{2}[$ from (ii) $]$
$\Rightarrow c^{2}=b^{2}$
$\Rightarrow c=\sqrt{b^{2}}$
$\Rightarrow c=b$
Substituting the value of c and a in eq. (iii), we get
Eccentricity, $e=\frac{c}{a}$
$=\frac{b}{b \sqrt{2}}$
$\therefore \mathrm{e}=\frac{1}{\sqrt{2}}$