Answer:
B.) (1, -5)
Given, equation:
[tex]\sf \dfrac{(x-1)^2}{25} -\dfrac{(y+5)^2}{49} =1[/tex]
Comparing it with standard formula of hyperbola:
[tex]\sf \dfrac{(x-h)^2}{a^2} -\dfrac{(y-k)^2}{b^2} =1[/tex]
where (h, k) is center
The given equation in standard form:
[tex]\sf \dfrac{(x-1)^2}{5^2} -\dfrac{(y-(-5))^2}{7^2} =1[/tex]
Determined:
(h, k) = (1, -5)
a = 5, b = 7
So, here (1, -5) is the center of the given hyperbola.
