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The directrices of the hyperbola $x^{2}-4(y-3)^{2}=16$ are

\[\begin{array}{1 1}(1)y=\pm\frac{8}{\sqrt{5}}&(2)x=\pm\frac{8}{\sqrt{5}}\\(3)y=\pm\frac{\sqrt{5}}{8}&(4)x=\pm\frac{\sqrt{5}}{8}\end{array}\]

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$x^2-4(y-3)^2=16$
$\large\frac{x^2}{16} -\frac{4 (y-3)^2}{16} $$=1$
$\large\frac{x^2}{16} -\frac{ (y-3)^2}{4} $$=1$
$\large\frac{(x-0)^2}{16} -\frac{4 (y-3)^2}{4} $$=1$
$\large\frac{X^2}{16} -\frac{Y^2}{4} $$=1$----(1)
Where $X= x-0$
$Y=y-3$
$a^2=16$
$b^2=4$
Eccentricity $e= \sqrt { 1+ \large\frac {b^2}{a^2}}$
$\qquad= \sqrt { 1+ \large\frac{4}{16}}$
$\qquad= \sqrt {\large\frac{20}{16}}$
$e= \large\frac{2 \sqrt 5}{4}$
$\quad= \large\frac{\sqrt 5}{2}$
$X= \pm \large\frac{a}{e}$
$x-0 = \pm \large\frac{4}{\sqrt 5}$
$x= \pm \large\frac{8}{\sqrt 5}$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
 
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