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The straight line $2x-y+c=0$ is a tangent to the ellipse $ 4x^{2}+8y^{2}=32$ if $c$ is

\[\begin{array}{1 1}(1)\pm 2\sqrt{3}&(2)\pm 6\\(3)36&(4)\pm4\end{array}\]

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The equation of the ellipse is
$4x^2+8y^2=32$
$\large\frac{4x^2}{32} + \frac{8y^2}{32} $$=1$
$\large\frac{x^2}{8} + \frac{y^2}{4} $$=1$-----(1)
The equation of the line is
$2x-y+c=0$
$y=2x+c$ ------(2)
The condition for the line $y=mx+c$ to touch the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{^2}$$=1$ is $c^2=a^2m^2+^2$
From (1) and (2)
$a^2=8, b^2=4,m=2,c=c$
$c^2=8 \times 2^2+4$
$c^2=32+4=36$
$c= \pm 6$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
 

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