Question

If the eccentricity of the ellipse $a{x}^2+4y^2=4a,(a<4)$ is $\frac{1}{\sqrt{2}}$, then its semi-minor axis is equal to

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

Answer 1

The correct option is B $\sqrt{2}$

Given equation of ellipse is
$a{x}^2+4y^2=4a...\left(i\right)$
We can write Eq.(i) as
$\frac{x^2}{4}+\frac{y^2}{a}=1$
Here $a^2=4;b^2=a$
$\because a<4$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{a}{4}}\Rightarrow \frac{1}{2}=1-\frac{a}{4}$
$\Rightarrow \frac{a}{4}=\frac{1}{2}\Rightarrow a=2$
$\therefore b^2=2\Rightarrow b=\sqrt{2}$
Hence, semi-minor axis is $b=\sqrt{2}$