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The eccentricity of the hyperbola $ 12y^{2}-4x^{2}-24x+48y-127=0$ is

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

1 Answer

$12y^2-4x^2-24x+48y-127=0$
$-4x^2-24x+12y^2+48y-127=0$
$-4(x^2+6x)+12(y^2+4y)-127=0$
$-4[(x+3)^2-3^2]+12(y+2)^2-127=0$
$-4(x+3)^2+36+12(y+2)^2-48-127=0$
$12(y+2)^2-4(x+3)^2=139$
$\large\frac{12(y+2)^2}{139} -\frac{4 (x+3)^2}{139}$$=1$
$\large\frac{(y+2)^2}{\Large\frac{139}{12} } - \frac{(x+3)^2}{\Large\frac{139}{4}}$$=1$
$a^2= \large\frac{139}{12}$
$b^2= \large\frac{139}{4}$
$e= \sqrt {\large\frac{a^2+^2}{a^2}}$
$\qquad= \sqrt { \large\frac{\Large\frac{139}{12}+\frac{139}{4}}{\frac{139}{12}}}$
$\qquad= \sqrt {\large\frac{139+417}{12} \times \frac{12}{139}}$
$\qquad= \sqrt {\large\frac{556}{139}}$
$\qquad=\sqrt 4$
$\qquad=2$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
 

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