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 $y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$
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 $$y-y_1=m(x-x_1)\implies y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$$
$$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$ $$ {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} \right)=0$$ $$ \frac{2}{9}(x- 3)- \frac{2}{4}(y- 2){dy\over dx}= 0 $$ $$ \frac{4(x- 3)}{9(y- 2)}= {dy\over dx} $$ $${dy\over dx}\biggr|_{(-2,{14\over3})}= -\frac{5}{6} $$
Thus the equation of the tangent is
$$(y-\frac{14}{3})=\frac{-5}{6}(x+2)$$
From the question on the Instant Tangent here, the equation of the tangent at $\big(-2, \frac {14}3\big)$ is $$\begin{align} \frac {(x-3)(-2-3)}9-\frac {(y-2)(\frac {14}3-2)}4&=1 \\5x+6y&=18\end{align}$$
The formula is the same as the one you quoted, except for the translation from origin to $(3,2)$.
