JEE MAIN - Mathematics (2006 - No. 29)
In the ellipse, the distance between its foci is $$6$$ and minor axis is $$8$$. Then its eccentricity is :
$${3 \over 5}$$
$${1 \over 2}$$
$${4 \over 5}$$
$${1 \over {\sqrt 5 }}$$
Обяснение
$$2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4$$
$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$
$$ \Rightarrow a{}^2 = 16 + 9 = 25$$
$$ \Rightarrow a = 5$$
$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$
$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$
$$ \Rightarrow a{}^2 = 16 + 9 = 25$$
$$ \Rightarrow a = 5$$
$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$