Given,
The center of the hyperbola is (-3, 2).
The length of the conjugate axis is 8 units.
The length of transverse axis is 12 units.
The center of the hyperbola is ,
[tex]\begin{gathered} (h,k)=(-3,2) \\ So,\text{ h = -3 and k =2} \end{gathered}[/tex]
As, the transverse axis is parallel to the x-axis then hyperbola must be opens on left and right.
The length of conjugate axis is calculated as:
[tex]\begin{gathered} \text{Length of conjugate axis = 2b} \\ 8=2b \\ b=4 \end{gathered}[/tex]
The length of transverse axis is calculated as:
[tex]\begin{gathered} \text{Length of transverse axis = 2a} \\ 12=2a \\ a=6 \end{gathered}[/tex]
The standard equation of hyperbola is,
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-b)^2}{b^2}=1[/tex]
Substituting the values then,
[tex]\begin{gathered} \frac{(x-(-3))^2}{6^2}-\frac{(y-2)^2}{4^2}=0 \\ \frac{(x+3)^2}{36^{}}-\frac{(y-2)^2}{16^{}}=0 \end{gathered}[/tex]
Hence, the equation of hyperbola is (x+3)^2/36-(y-2)^2/16=0
