If $S$ and $S^{\prime}$ are the foci of an ellipse, $B$ is one end of the minor axis and $\angle S B S^{\prime}=90^{\circ}$, then the eccentricity of that ellipse is

Question

Aicharya · Accepted Answer

$S$ and $S^1$ are the foci of an ellipse $\therefore$ Coordinate of foci $=( \pm a e, 0)$

Slope of $S B=\frac{b}{a e}$ Slope of $S^1 B=\frac{-b}{a e}$ Now, $\angle S B S^{\prime}=90$ So, slope of $S B \times$ slope of $S^{\prime} B=-1$ $ \begin{array}{ll} \Rightarrow & \frac{b}{a e} \times\left(-\frac{b}{a e}\right)=-1 \\ \Rightarrow & \frac{b^2}{a^2 e^2}=1 \Rightarrow b^2=a^2 e^2 \end{array} $ We know that, $ \begin{aligned} b^2 & =a^2\left(1-e^2\right) \\ \Rightarrow \quad a^2 e^2 & =a^2\left(1-e^2\right) \Rightarrow e^2=1-e^2 \\ \Rightarrow \quad 2 e^2 & =1 \Rightarrow e=\frac{1}{\sqrt{2}} \end{aligned} $

If $S$ and $S^{\prime}$ are the foci of an ellipse, $B$ is one end of the minor axis and $\angle S B…

Solution