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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If the angle between the straight line joining foci and the ends of the minor axis of the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is $90^{\large\circ}$ .Find its eccentricity?

$\begin{array}{1 1}(a)\;\large\frac{1}{\sqrt 2}\\(b)\;\large\frac{1}{\sqrt 5}\\(c)\;\large\frac{3}{\sqrt 2}\\(d)\;\large\frac{5}{\sqrt 2}\end{array}$

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Equation of ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}=$$1$(a > b)
Slope of BS=$\large\frac{-b}{ae}$$=m_1$
Slope of BS'=$\large\frac{b}{ae}$$=m_2$
Angle between BS and BS' is $90^{\large\circ}$ hence
$m_1m_2=-1$
$b^2=a^2e^2$
$a^2(1-e^2)=a^2e^2$
$b^2=a^2(1-e^2)$
$2e^2=1$
$e=\large\frac{1}{\sqrt 2}$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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