question_answer

If the angle between the straight lines joining the foci to an extremity of minor axis in an ellipse be \[90{}^\circ \]; then the eccentricity of the ellipse is

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

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

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

D) \[\frac{1}{3}\]

Correct Answer: C

Solution :

[c] From standard equation of

ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], the

Co-ordinates of foci are

\[S(ae,0)\] And \[S'(-ae,0).\]

Co-ordinate of an extremity of the minor axis is B (0, b),

Now slope of straight line

\[BS=\frac{b-a}{0-ae}=\frac{b}{-ae}={{m}_{1}}\]

And slope of straight line

\[BS'=\frac{b-a}{0-(-ae)}=\frac{b}{ae}={{m}_{2}}\]

\[\because SB\bot BS',\] so \[{{m}_{1}}.{{m}_{2}}=-1\] or

\[\frac{b}{-ae}\times \frac{b}{ae}=-1;\] or \[\frac{{{b}^{2}}}{{{a}^{2}}}={{e}^{2}}\]

But

\[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}});\] or \[\frac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}};\] or \[{{e}^{2}}=1-{{e}^{2}}.\]

Or \[2{{e}^{2}}=1;\] or \[e=\frac{1}{\sqrt{2}}.\]