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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the eccentricity of ellipse $\large\frac{x^2}{64}+\frac{y^2}{100}=1$?

$\begin{array}{1 1}(a)\;\large\frac{3}{4}\\(b)\;\large\frac{3}{5}\\(c)\;\large\frac{3}{8}\\(d)\;\large\frac{5}{3}\end{array}$

1 Answer

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$a^2=64,b^2=100$
$a^2=b^2(1-e^2)$
$64=100(1-e^2)$
$\large\frac{64}{100}$$-1=-e^2$
$\large\frac{36}{100}$$=e^2$
$e=\large\frac{3}{5}$
Hence (b) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
 

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