question_answer

If the lines joining the foci of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}\,+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\] where\[a>b,\]and an extremly of its minor axis are inclined at an angle\[60{}^\circ ,\]then the eccentricity of the ellipse is

A)  \[-\frac{\sqrt{3}}{2}\]

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

C)  \[\frac{\sqrt{5}}{2}\]                    

D)         \[\frac{\sqrt{7}}{3}\]

E)  \[\sqrt{3}\]

Correct Answer: B

Solution :

In\[\Delta CBF,\tan 30{}^\circ =\frac{FC}{b}\] \[FC=\frac{b}{\sqrt{3}}\] \[\Rightarrow \]\[ae=\frac{b}{\sqrt{3}}\]\[\Rightarrow \]\[{{a}^{2}}{{e}^{2}}=\frac{1}{2}[{{a}^{2}}(1-{{e}^{2}})]\] \[\Rightarrow \]\[4{{e}^{2}}=1\]\[\Rightarrow \]\[e=\frac{1}{2}\]