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.

Your hyperbola is $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= 1$ and you want to find the tangent line to it "at (-2, 14/3)".  The first thing I would do is check to make sure that point is on the hyperbola.  With x= -2, $(x- 3)^2= 25$ and $\frac{(x- 3)^2}{9}= \frac{25}{9}$.  With $y= \frac{14}{3}$, $(y- 2)^2= \frac{64}{9}$ and $\frac{y- 14/3}{4}= \frac{16}{9}$.  Yes, a bit to my surprise, $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= \frac{25}{9}- \frac{16}{9}= \frac{9}{9}= 1$ and the point is on the hyperbola!
Now, do you know what a "tangent line" is and how to find the slope of a tangent line without simply plugging into formulas?  Using "implicit differentiation" to differentiate both sides with respect to x, we have $\frac{2}{9}(x- 3)- \frac{1}{2}(y- 2)y'= 0$.  In particular, at x= -2, y= 14/3, $\frac{2}{9}(-5)- \frac{1}{2}\frac{8}{3}y'= -\frac{10}{9}-\frac{4}{3}y'= 0$ so $y'= -\left(\frac{3}{4}\right)\left(\frac{10}{9}\right)= -\frac{5}{6}$.  The tangent line is $y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$
