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

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

$9x^2+4y^2-6x+4y-1=0$
=> $9 (x - \large\frac{1}{3})^2+4 (y+1)^2=1$