Conic Sections Question 336
Question: An ellipse has OB as semi minor axis, F and F¢ its foci and the angle FBF¢ is a right angle. Then the eccentricity of the ellipse is
[AIEEE 2005]
Options:
A) $ \frac{1}{4} $
B) $ \frac{1}{\sqrt{3}} $
C) $ \frac{1}{\sqrt{2}} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \angle {F}‘BF=90{}^\circ $ , $ {F}‘B\bot FB $
i.e., slope of $ ({F}‘B) $ ´ Slope of $ ({F}‘B)=-1 $
therefore $ \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} $ , $ 2e^{2}=1 $ , $ e^{2}=\frac{1}{2} $ , $ e=\frac{1}{\sqrt{2}} $ .
BETA
