Question

Let $S$ and $S^{\prime }$ be the foci of an ellipse and $B$ be one end of its minor axis. If $SB{S}^{\prime }$ is an isosceles right-angled triangle then the eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{3}$

Answer 1

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

In an ellipse, lenght of focus from vertex is $ae$, where a is lenght of semi-major axis and e is eccentricity.

=> $OS=ae$

lenght of semi-minor axis = $OB=b$

Now, $tan45=\frac{OS}{OB}=1$

=> $\frac{ae}{b}=1$

=>$e=\frac{b}{a}$

Now, eccentriciity is given by $e=\sqrt{1-\frac{b^2}{a^2}}$

=> $e^2=1-e^2$

=> $e=\frac{1}{\sqrt{2}}$