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10-2. Parabola, Ellipse, Hyperbola
hard
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
A
$\frac{1}{4}$
B
$\frac{1}{{\sqrt 3 }}$
C
$\frac{1}{{\sqrt 2 }}$
D
$\frac{1}{2}$
(AIEEE-2005)
Solution
(c) $\angle F'BF = 90^\circ $, $F'B \bot FB$
$i.e.$, slope of $(F'B)$ $\times$ Slope of $(FB) = – 1$
==> $\frac{b}{{ae}} \times \frac{b}{{ – ae}} = – 1$, ${b^2} = {a^2}{e^2}$…..$(i)$
We know that $e = \sqrt {1 – \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 – \frac{{{a^2}{e^2}}}{{{a^2}}}} = \sqrt {1 – {e^2}} $
${e^2} = 1 – {e^2}$, $2{e^2} = 1$,
${e^2} = \frac{1}{2}$, $e = \frac{1}{{\sqrt 2 }}$.
Standard 11
Mathematics
