Conic Sections Question 286
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
Options:
A) $ \frac{1}{\sqrt{2}} $
B) $ \frac{1}{2} $
C) $ \frac{1}{4} $
D) $ \frac{1}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \because \angle FBF’=90{}^\circ \Rightarrow FB^{2}+F’B^{2}=FF{{’}^{2}} $
$ \therefore {{( \sqrt{a^{2}e^{2}+b^{2}} )}^{2}}+{{( \sqrt{a^{2}e^{2}+b^{2}} )}^{2}}={{(2ae)}^{2}} $
$ \Rightarrow 2(a^{2}e^{2}+b^{2})=4a^{2}e^{2}\Rightarrow e^{2}=\frac{b^{2}}{a^{2}} $ …… (i) Also, $ e^{2}=1-b^{2}/a^{2}=1-e^{2} $ (By using equation (i))
$ \Rightarrow 2e^{2}=1\Rightarrow e=\frac{1}{\sqrt{2}}. $
BETA
