Question

BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

B.) (1, -5)

Given, equation:

$\sf \dfrac{(x-1)^2}{25} -\dfrac{(y+5)^2}{49} =1$

Comparing it with standard formula of hyperbola:

$\sf \dfrac{(x-h)^2}{a^2} -\dfrac{(y-k)^2}{b^2} =1$

where (h, k) is center

The given equation in standard form:

$\sf \dfrac{(x-1)^2}{5^2} -\dfrac{(y-(-5))^2}{7^2} =1$

Determined:

(h, k) = (1, -5)

a = 5, b = 7

So, here (1, -5) is the center of the given hyperbola.