A) \[\frac{1}{4}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[\angle {F}'BF=90{}^\circ \], \[{F}'B\bot FB\] i.e., slope of \[({F}'B)\] ´ Slope of \[({F}'B)=-1\] Þ \[\frac{b}{ae}\times \frac{b}{-ae}=-1\], \[{{b}^{2}}={{a}^{2}}{{e}^{2}}\] ?..(i) We know that \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{{{a}^{2}}{{e}^{2}}}{{{a}^{2}}}}=\sqrt{1-{{e}^{2}}}\] \[{{e}^{2}}=1-{{e}^{2}}\], \[2{{e}^{2}}=1\], \[{{e}^{2}}=\frac{1}{2}\], \[e=\frac{1}{\sqrt{2}}\].You need to login to perform this action.
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