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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!!}}}

User Jturolla
by
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1 Answer

4 votes

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).

User Akshay Bhange
by
8.6k points
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