The focus of the curve $y^2+4x-6y+13=0$ is.

The given equation of curve is

$y^2+4x-6y+13=0$

which can be written as :

$\begin{array}{l}{\mathrm{y}}^2-6\mathrm{y}+9+4\mathrm{x}+4=0\\ \Rightarrow \left({\mathrm{y}}^2-6\mathrm{y}+9\right)=-4(\mathrm{x}+1)\\ \Rightarrow (\mathrm{y}-3{)}^2=-4(\mathrm{x}+1)\end{array}$

Put $\mathrm{Y}=\mathrm{y}-3\text{ and }\mathrm{X}=\mathrm{x}+1$

On comparing ${\mathrm{Y}}^2=4\mathrm{aX}$

Length of focus from vertex, $a=-1$

At focus X = a and Y = 0

$\begin{array}{l}\Rightarrow x+1=-1\Rightarrow x=-2\\ y-3=0\Rightarrow y=3\end{array}$

$\therefore$ Focus is(-2,3).