Question

If S and S are two foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and B is an end of the minor axis such that $△BS{S}^{\prime }$ is equilateral, then write the eccentricity of the ellipse.

Answer 1

The given equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow$ Now, $△BS{S}^{\prime }$ is equilateral [given]
BS= SS'= BS'
$\Rightarrow (BS{)}^2=(S{S}^{\prime }{)}^2=(B{S}^{\prime }{)}^2$
$\therefore (BS{)}^2=(s{s}^{\prime }{)}^2$
$\Rightarrow (0-ae{)}^2+(b-0{)}^2$
$=(ae+ae{)}^2+(0-0{)}^2$
$\Rightarrow (ae{)}^2+b^2=(2ae{)}^2$
$\Rightarrow (ae{)}^2+b^2=4(ae{)}^2$
$\Rightarrow b^2=4(ae{)}^2-(ae{)}^2$
$\Rightarrow b^2=3a^2{e}^2$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow 3a^2{e}^2=a^2(1-e^2)$
$\Rightarrow 3e^2=1-e^2$
$\Rightarrow 3e^2+e^2=1$
$\Rightarrow 4e^2=1$
$\Rightarrow e^2=\frac{1}{4}$
$\Rightarrow 3e^2=1-e^2$
$\Rightarrow 3e^2+e^2=1$
$\Rightarrow 4e^2=1$
$\Rightarrow e^2=\frac{1}{4}$
$\Rightarrow e=\frac{1}{2}$