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The angle between the asymptotes to the hyperbola $\large\frac{x^{2}}{16}-\frac{y^{2}}{9}=$$1 $ is

\[\begin{array}{1 1}(1)\pi-2\tan^{-1}\left(\begin{array}{c} 3 \\ 4 \end{array}\right)&(2)\pi-2\tan^{-1}\left(\begin{array}{c} 4 \\ 3 \end{array}\right)\\(3)2\tan^{-1}\frac{3}{4}&(4)2\tan^{-1}\left(\begin{array}{c} 4 \\ 3 \end{array}\right)\end{array}\]

1 Answer

$\large\frac{x^2}{16} -\frac{y^2}{9} $$=1$
$a^2= 167,b^2=9 => a= 4,b=3$
Angle between the asymptotes $=2 \tan^{-1} \bigg( \large\frac{b}{a} \bigg)$
$\qquad= 2 \tan ^{-1} \bigg( \large\frac{3}{4} \bigg)$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
 

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