LR of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a < b)$ is half of its major axis. If its eccentricity is $e = \frac{1}{\sqrt{k}}$ , then value of $k$ is

Open in App

Solution

given that $L R = b$
$\Rightarrow \frac{2 a^{2}}{b} = b$
$\Rightarrow \frac{a^{2}}{b^{2}} = \frac{1}{2}$
$e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \frac{1}{\sqrt{2}}$
Ans: $2$

Suggest Corrections

Similar questions

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

(\frac{x^{2}}{a^{2}}) + (\frac{y^{2}}{b^{2}}) = 1

e

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

e^{'}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b)

e

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

(a < b)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Join BYJU'S Learning Program