Question

The eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$whose latus rectum is half of its major axis, is .

• A. $e=\frac{1}{\sqrt{2}}$
• B. $e=\frac{1}{2}$
• C. $e=\frac{1}{2\sqrt{2}}$
• D. $e=\frac{1}{4}$

Answer 1

The correct option is A $e=\frac{1}{\sqrt{2}}$

$\mathrm{Let}a>b.\mathrm{Then},\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{is}$$\frac{2b^2}{a}$$\mathrm{and}\mathrm{the}\mathrm{semi}-\mathrm{major}\mathrm{axis}\mathrm{is}a.$
$\mathrm{Given},\frac{2b^2}{a}=a\mathrm{or}2b^2=a^2$
Also for the ellipse,
$b^2=a^2(1-e^2)b^2=2b^2(1-e^2)\mathrm{On}\mathrm{solving},e=\frac{1}{\sqrt{2}}$