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$16x^{2}-3y^{2}-32x-12y-44=0 $ represents

\[\begin{array}{1 1}(1)an\; ellipse &(2)a\; circle\\(3)a\; parabola&(4)a\; hyperbola\end{array}\]

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$Ax^2+Bxy+Cy^2+2Gx+2Fy+D=0$------(1)
Represent a parabola if $B^2-4AC=0$
an ellipse if $B^2-4AC < 0$ a hyperbola if $ B^2-4AC > 0$.
The given equation is
$16x^2-3y^2-32x-12y-44=0$-----(2)
Comparing (1) and (2)
$A= 16,B =0,C=-3,2G=-32,2F=-12,D=-44$
$B^2-4AC=0^2 -4 (16) (-3)= +4 (48) > 0$
It is hyperbola
Hence 4 is the correct answer.
answered May 15, 2014 by meena.p
 

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