question_answer

In an ellipse the angle between the lines joining the foci with the positive end of minor axis is a right angle, the eccentricity of the ellipse is

A) \[\frac{1}{\sqrt{2}}\]                     

B)         \[\frac{1}{\sqrt{3}}\]                    

C) \[\sqrt{2}\]                        

D)         \[\sqrt{3}\]

Correct Answer: A

Solution :

Let \[S(ae,0)\]and \[S(-ae,0)\]are foci of an ellipse and \[B(0,b)\]is the positive end of minor axis. \[\therefore \]  Slope of \[SB=\frac{b-0}{0-ae}=-\frac{b}{ae}\] and slope of SB\[=\frac{b}{ae}.\] Since SB and SB are perpendicular to each other \[\therefore \]  \[\left( -\frac{b}{ae} \right)\left( \frac{b}{ae} \right)=-1\] \[\Rightarrow \]               \[{{b}^{2}}={{a}^{2}}{{e}^{2}}\] But         \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[\therefore \]  \[{{a}^{2}}-{{a}^{2}}{{e}^{2}}={{a}^{2}}{{e}^{2}}\] \[\Rightarrow \]               \[{{a}^{2}}=2{{a}^{2}}{{e}^{2}}\] \[\Rightarrow \]               \[e=\frac{1}{\sqrt{2}}\]