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The length of the latus rectum of the parabola $y^{2}-4x+4y+8=0$ is

\[\begin{array}{1 1}(1)8&(2)6\\(3)4&(4)2\end{array}\]

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$y^2-4x+4y+8=0$
$y^2+4y-4x+8=0$
$(y+2)^2-2^2-4x+8=0$
$(y+2)^2-4-4x+8=0$
$(y+2)^2 =4x-4$
$(y+2)^2=4 (x+1)$
$a=1$
So $4a=4$
Hence 3 is the correct answer
answered May 15, 2014 by meena.p
 

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