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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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For the hyperbola $\large\frac{x^2}{36}-\frac{y^2}{16}$$=1$ find the ecentricity?

$\begin{array}{1 1}(a)\;\large\frac{\sqrt{52}}{6}\\(b)\;\large\frac{\sqrt{50}}{6}\\(c)\;\large\frac{\sqrt{32}}{6}\\(d)\;\large\frac{\sqrt{22}}{6}\end{array}$

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1 Answer

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$e=\sqrt{1+\Large\frac{b^2}{a^2}}$
$e=\sqrt{1+\Large\frac{16}{36}}$
$e=\sqrt{\Large\frac{52}{6}}$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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