<s>[INST] 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.
 [/INST] $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$
$$ {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4}  \right)=0$$
$$ \frac{2}{9}(x- 3)- \frac{2}{4}(y- 2){dy\over dx}= 0 $$
$$ \frac{4(x- 3)}{9(y- 2)}= {dy\over dx} $$
$${dy\over dx}\biggr|_{(-2,{14\over3})}= -\frac{5}{6} $$
Thus the equation of the tangent is 
$$(y-\frac{14}{3})=\frac{-5}{6}(x+2)$$
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