The general equation of a hyperbola is represented as: [tex]\mathbf{\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1}[/tex]
The equation of the hyperbola is: [tex]\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}[/tex]
Recall that:
[tex]\mathbf{\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1}[/tex]
The hyperbola lies on [tex]\mathbf{y = -3}[/tex]
This means that: [tex]\mathbf{k =y}[/tex]
So, we have:
[tex]\mathbf{k =-3}[/tex]
The length on the y-axis is 6.
So:
[tex]\mathbf{a = \frac 62}[/tex]
[tex]\mathbf{a = 3}[/tex]
Similarly, the hyperbola lies on [tex]\mathbf{x = 2}[/tex]
This means that: [tex]\mathbf{h =x}[/tex]
So, we have:
[tex]\mathbf{h =2}[/tex]
The length on the x-axis is 8.
So:
[tex]\mathbf{b = \frac 82}[/tex]
[tex]\mathbf{b = 4}[/tex]
At this point, we have:
[tex]\mathbf{k =-3}[/tex]
[tex]\mathbf{a = 3}[/tex]
[tex]\mathbf{h =2}[/tex]
[tex]\mathbf{b = 4}[/tex]
Substitute these values in [tex]\mathbf{\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1}[/tex]
[tex]\mathbf{\frac{(x - 2)^2}{3^2} - \frac{(y -- 3)^2}{4^2} = 1}[/tex]
[tex]\mathbf{\frac{(x - 2)^2}{3^2} - \frac{(y + 3)^2}{4^2} = 1}[/tex]
Evaluate the exponents
[tex]\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}[/tex]
Hence, the equation of the hyperbola is: [tex]\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}[/tex]
Read more about equations of hyperbola at:
https://brainly.com/question/12919612
