# Question #58fdd

Apr 13, 2017

$\left(\pm \frac{\sqrt{3}}{2} , 0\right) .$

#### Explanation:

${x}^{2} = 1 - 4 {y}^{2} \Rightarrow {x}^{2} + 4 {y}^{2} = 1 , \mathmr{and} , {x}^{2} + {y}^{2} / \left(\frac{1}{4}\right) = 1$

Comparing with the Standard Eqn. of Ellipse, i.e.,

${x}^{2} / {a}^{2} + {t}^{2} / {b}^{2} = 1 ,$ we have, ${a}^{2} = 1 , {b}^{2} = \frac{1}{4} , s o , a > b .$

Knowing that the Eccentricity $e$ is given by,

${b}^{2} = {a}^{2} \left(1 - {e}^{2}\right) ,$ we get, $\frac{1}{4} = 1 \left(1 - {e}^{2}\right)$

$\therefore e = \frac{\sqrt{3}}{2.}$

Therefore, the Focii are, $\left(\pm a e , 0\right) = \left(\pm \frac{\sqrt{3}}{2} , 0\right) .$

Enjoy Maths.!