The equation of the hyperbola that has a center at (1,2) , a focus at (-4 , 2) , and a vertex at (-3 , 2 ) , is \frac((x-C)^2)(A^2)-\frac((y

The equation of the hyperbola that has a center at (1,2) , a focus at (-4 , 2) , and a vertex at (-3 , 2 ) , is ((x-C)^2)/(A^2)-((y-D)^2)/(B^2)=1

$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$

center at (1,2):   

 $$\small{\text{$
\begin{array}{rcl}
x-C &=& 0 \\ 1-C &=& 0 \\ C &=& 1 \\\\
y-D &=& 0 \\ 2-D &=& 0 \\ D &=& 2
\end{array}
$}}$$

vertex at (-3 , 2 ):

$$\small{\text{$
\begin{array}{l}
(-3 , 2 ) - (1,2)_{\mathrm{center}} = (-4,0) =(\pm A ,0)\\
A = \pm 4
\end{array}
$}}$$

focus at (-4 , 2):

$$\small{\text{$
\begin{array}{l}
(-4 , 2 ) - (1,2)_{\mathrm{center}} = (-5,0) =(\pm \sqrt{A^2+B^2} ,0)\end{array}
$}}\\
\begin{array}{rcl}
A^2+B^2 &=& (-5)^2 = 25\\
(\pm 4)^2+B^2 &=& 25\\
B^2 &=& 25-16 = 9\\
B &=& \pm 3
\end{array}
$}}$$

A = $$\small{\text{$\pm 4$}}$$

B = $$\small{\text{$\pm 3$}}$$

C = 1

D = 2

$$\dfrac{(x-1)^2}{(\pm 4)^2}-\dfrac{(y-2)^2}{(\pm 3)^2}=1$$

$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$

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

$$\\4^2+B^2=5^2\\
B=3$$

$$\\\dfrac{(x-1)^2}{4^2}-\dfrac{(y-2)^2}{3^2}=1\\\\\\
\dfrac{(x-1)^2}{16}-\dfrac{(y-2)^2}{9}=1$$

Very nice, heureka and Melody.......

Thanks Chris