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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Equation of the director circle of the ellipse $x^2+2y^2+2x-12y+15=0$ is

$\begin{array}{1 1}(A)\;x^2+y^2+2x-6y+4=0 \\(B)\;x^2+y^2+2x-12y+4=0 \\(C)\;x^2+y^2+2x-6y-4=0 \\(D)\;x^2+y^2+2x-12y-4=0 \end{array}$

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1 Answer

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=> $ \large\frac{(x+1)^2}{4}+\frac{(y-3)^2}{2}$$=1$
$\therefore $ Equation of its director circle is
=> $x^2+y^2+2x-6y+4=0$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p

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