$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$
The equation of the hyperbola that has a center at (1,2) ,
$$\dfrac{(x-1)^2}{A^2}-\dfrac{(y-2)^2}{B^2}=1$$ and $$C^2=A^2+B^2$$
a focus at (-4 , 2) , and a vertex at (-3 , 2 )
The foci are at $$(h\pm C,k)$$ so 1-C=-4 C=5
The vertices are at $$(h\pm A, k)$$ 1-A=-3 A=4
$$\\4^2+B^2=5^2\\
B=3$$
$$\\\dfrac{(x-1)^2}{4^2}-\dfrac{(y-2)^2}{3^2}=1\\\\\\
\dfrac{(x-1)^2}{16}-\dfrac{(y-2)^2}{9}=1$$
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