Hii! I need some help, would anyone mind helping me?
Find the center of this hyperbola.
[tex]\boldsymbol{x^2-25y^2-14x+100y-76=0}[/tex]
[tex]\stackrel\bigstar{\boldsymbol{\underbrace{Thankyou!!}}}[/tex]

Respuesta :

Step-by-step explanation:

[tex] {x}^{2} - 14x - 25 {y}^{2} + 100y = 76[/tex]

Factor by grouping,

[tex]( {x}^{2} - 14x) - (25 {y}^{2} - 100y) = 76[/tex]

Complete the square, with the x variables,

[tex]( {x}^{2} - 14x + 49) - (25 {y}^{2} - 100y) = 125[/tex]

Factor out 25 for the y variables

[tex]( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y) = 125[/tex]

Complete the square

[tex]( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y + 4) = 25[/tex]

Simplify the perfect square trinomial

[tex](x - 7) {}^{2} - 25(y - 2) {}^{2} = 25[/tex]

Make the right side be 1 so divide everything by 25.

[tex] \frac{(x - 7) {}^{2} }{25} - (y - 2) {}^{2} = 1[/tex]

Here our center is (7,2).