Question

Hii! I need some help, would anyone mind helping me?

Find the center of this hyperbola.

`\boldsymbol{x^2-25y^2-14x+100y-76=0}`

`\stackrel\bigstar{\boldsymbol{\underbrace{Thankyou!!}}}`

Answer 1

Explanation:

`{x}^(2) - 14x - 25 {y}^(2) + 100y = 76`

Factor by grouping,

`( {x}^(2) - 14x) - (25 {y}^(2) - 100y) = 76`

Complete the square, with the x variables,

`( {x}^(2) - 14x + 49) - (25 {y}^(2) - 100y) = 125`

Factor out 25 for the y variables

`( {x}^(2) - 14x + 49) - 25( {y}^(2) - 4y) = 125`

Complete the square

`( {x}^(2) - 14x + 49) - 25( {y}^(2) - 4y + 4) = 25`

Simplify the perfect square trinomial

`(x - 7) {}^(2) - 25(y - 2) {}^(2) = 25`

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

`\frac{(x - 7) {}^(2) }{25} - (y - 2) {}^(2) = 1`

Here our center is (7,2).