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Question

Mathematics Question on Ellipse

BB is an extremity of the minor axis of an ellipse whose foci are SS and SS'. If ?SBS?SBS' is a right angle, then the eccentricity of the ellipse is

A

12\frac{1}{2}

B

12\frac{1}{\sqrt{2}}

C

23\frac{2}{3}

D

13\frac{1}{3}

Answer

12\frac{1}{\sqrt{2}}

Explanation

Solution

Slope of SB,m1=b00ae=baeSB, m_1 = \frac{b - 0}{0 - ae} =-\frac{b}{ae}
and slope of SB,m2=b00(ae)=baeS^{\prime} B, m_{2}=\frac{b-0}{0-(-a e)}=\frac{b}{a e}
Since, SBS\angle S B S^{\prime} is a right angle.
m1m2=1\therefore m_{1} m_{2} =-1
bae×bae=1\Rightarrow \frac{-b}{a e} \times \frac{b}{a e} =-1
b2=a2e2\Rightarrow b^{2} =a^{2} e^{2}
b2a2=e2\Rightarrow \frac{b^{2}}{a^{2}} =e^{2}
1e2=e2\Rightarrow 1-e^{2} =e^{2}
2e2=1\Rightarrow 2 e^{2} =1
e2=12\Rightarrow e^{2}=\frac{1}{2}
e=±12\Rightarrow e=\pm \frac{1}{\sqrt{2}}
e=12\Rightarrow e=\frac{1}{\sqrt{2}}
or e=12 e=-\frac{1}{\sqrt{2}}