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The line $5x-2y+4k=0 $ is a tangent to $4x^{2}-y^{2}=36$ then $k$ is

\[\begin{array}{1 1}(1)\frac{4}{9}&(2)\frac{2}{3}\\(3)\frac{9}{4}&(4)\frac{81}{16}\end{array}\]

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1 Answer

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The equation of the hyperbola is
$4x^2-y^2=36$
$\large\frac{4x^2}{36} -\frac{y^2}{36}$$=1$
$\large\frac{4x^2}{9} -\frac{y^2}{36}$$=1$----(1)
$a^2=9;$
$b^2=36$
The equation of the line is
$5x-2y+4k=0$
$2y=5x+4k$
$y= \large\frac{5}{2}$$+2k$----(2)
From equation (2) $m= \large\frac{5}{2}$$c= 2k$
$c^2= a^2m^2-b^2$
$(2k)^2 = 9 \bigg( \large\frac{5}{2} \bigg) ^2$$ -36$
$4k^2 = 9 \times \large\frac{25}{4} $$-36$
$4k^2 = \large\frac{225 -144}{4}$
$k^2=\large\frac{81}{16} $$=> k= \large\frac{9}{4}$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
 

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