Conic Sections Question 445
If the angle between the lines joining the end points of minor axis of an ellipse with its foci is $ \frac{x^{2}}{25}+\frac{y^{2}}{25}=1 $ , then the eccentricity of the ellipse is
[IIT Screening 1997; Pb. CET 2001; DCE 2002]
Options:
1/2
B) $ 1/\sqrt{2} $
C) $ \sqrt{3}/2 $
D) $ 1/2\sqrt{2} $
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Answer:
Correct Answer: B
Solution:
Since $ \angle FBF’=\frac{\pi }{2} $ (Given)
$ \therefore $ $ \angle FBC=\angle F’BC=\pi /4 $
$ CB=CF\Rightarrow b=ae $
therefore $ b^{2}=a^{2}(1-e^{2}) $
therefore $ a^{2}(1-e^{2}) \neq a^{2}e^{2} $
therefore $ 1-e^{2} \neq e^{2} $
therefore $ 2e^{2}\neq1 $
therefore $ e=1/\sqrt{2} $ .
BETA
