How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$  is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$. 
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$ but cant figure how to apply those here since the centre is not at origin.

given hyperbola is $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$ now differentiate the equation with respect to x to get the slope of tangent at that point $$\implies {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} = 1 \right)\implies \frac{2}{9}(x- 3)- \frac{1}{2}(y- 2){dy\over dx}= 0 \\ \implies{dy\over dx}= -\frac{5}{6} \ \ \ at\ (x_1,y_1)=(-2,{14\over3})$$ and the tangent line is $$y-y_1=m(x-x_1)\implies y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$$
