Question

find the center of this hyperbola.
$x^2 - 25y^2 - 14x + 100y - 76 = 0$
center : ( ? , )

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \((x^2 - 14x) + (-25y^2 + 100y) = 76\).

Step2: Factor out coefficients of squared terms

Factor out \(-25\) from the \(y\)-terms: \((x^2 - 14x) - 25(y^2 - 4y) = 76\).

Step3: Complete the square for x

For \(x^2 - 14x\), take half of \(-14\) (which is \(-7\)), square it (\(49\)), and add it: \(x^2 - 14x + 49 = (x - 7)^2\).

Step4: Complete the square for y

For \(y^2 - 4y\), take half of \(-4\) (which is \(-2\)), square it (\(4\)), and add it inside the \(y\)-group. But since there's a factor of \(-25\) outside, we actually add \(-25\times4 = -100\) to the left side.

Step5: Adjust the equation

Add \(49\) (from x) and \(-100\) (from y) to the right side: \((x - 7)^2 - 25(y - 2)^2 = 76 + 49 - 100\).
Calculate the right side: \(76 + 49 - 100 = 25\). So the equation becomes \(\frac{(x - 7)^2}{25} - \frac{(y - 2)^2}{1} = 1\).
The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\) (or similar for vertical transverse axis), where \((h, k)\) is the center.
From our equation, \(h = 7\) and \(k = 2\).

Answer:

\((7, 2)\)