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
where A =____
B =_____
C =______
D =______
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
$$\dfrac{(x-1)^2}{(\pm 4)^2}-\dfrac{(y-2)^2}{(\pm 3)^2}=1$$
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
$$\\4^2+B^2=5^2\\
B=3$$
$$\\\dfrac{(x-1)^2}{4^2}-\dfrac{(y-2)^2}{3^2}=1\\\\\\
\dfrac{(x-1)^2}{16}-\dfrac{(y-2)^2}{9}=1$$