Question

Eccentricity of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that the line joining the foci subtends a right angle only at two points on ellipse is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

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

Given equation of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$

Slope of $BS=\frac{-b}{ae}=m_1$

Slope of $B{S}^{{}^{\prime }}=\frac{b}{ae}=m_2$

Angle between $BS$ and $B{S}^{{}^{\prime }}$ is ${90}^0$

We know that

$slopeof{m}_1\times slopeof{m}_2=-1$

Then,

$m_1{m}_2=-1$

$\frac{-b}{ae}\times \frac{b}{ae}=-1$

$\frac{b^2}{{\left(ae\right)}^2}=1$

$b^2=a^2{e}^2$

$a^2(1-e^2)=a^2{e}^2$ Since $b^2=a^2(1-e^2)$

$(1-e^2)=e^2$

$1=2e^2$

$e^2=\frac{1}{2}$

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

Hence, option (B) is correct answer.