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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

The lengths of the axes of the conic $9x^2+4y^2-6x+4y+1=0$ are

$\begin{array}{1 1}(A)\;(\large\frac{1}{2},9) \\(B)\;(3,\frac{2}{5}) \\(C)\;(3,\frac{2}{5}) \\(D)\;(1, \frac{2}{3}) \end{array}$

1 Answer

$9x^2+4y^2-6x+4y-1=0$
=> $9 (x - \large\frac{1}{3})^2+4 (y+1)^2=1$
=> $\large\frac{(x- \Large\frac{1}{3})^2}{(\Large\frac{1}{3})^2}+\frac{(y+1)^2}{(\Large\frac{1}{2})^2}$$=1$
$\therefore $ which is ellipse with length of axis $(\large\frac{2}{3},1)$
Hence D is the correct answer.
answered Apr 10, 2014 by meena.p
 
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