Question

Identify the standard form of the equation by Completing the square for 4x^2 - 9y^2 - 8x +36y - 68=0

Answer 1

`((x-1)^2)/(9)-((y-2)^2)/(4)=1`

Explanation:

`4x^2 - 9y^2 - 8x +36y - 68=0\\4x^2-8x-9y^2+36y=68\\4(x^2-2x)-9(y^2-4y)=68\\4(x^2-2x+1)-9(y^2-4y+4)=68+4(1)-9(4)\\4(x-1)^2-9(y-2)^2=36\\((x-1)^2)/(9)-((y-2)^2)/(4)=1`

This is a hyperbola