Answer:
(x+1)²/6 - (y+3)²/12 = 1
Explanation:
The standard form of writing the equation of an hyperbola is expressed as;
(x-h)²/a² - (y-b)²/b² = 1 where (h,k) is the centre of the hyperbola.
Given the equation:
6x²-3y²+12x-18y-3 = 0
We are to convert it to the standard form of writing the equation of a hyperbola.
Collecting the like terms will give;
(6x²+12x)-(3y²+18y)-3 = 0
Divide through by 3
(2x²+4x)-(y²+6y)-1 = 0
Completing the square of the equation in parenthesis and adding the constants to the other side of the equation:
(2x²+4x)-(y²+6y+(6/2)²)-1 = 0+(6/2)²
(2x²+4x)-(y²+6y+9)-1 = 9
2x²+4x -{(y+3)²} = 9+1
2x²+4x - (y+3)² = 10
Divide through by 2
x²+2x -(y+3)²/2 = 5
(x²+2x+(2/2)²)-(y+3)²/2 = 5+(2/2)²
(x²+2x+1) - (y+3)²/2 = 5+1
(x+1)²-(y+3)²/2 = 6
Divide through by 6
(x+1)²/6 - (y+3)²/12 = 1
The resulting equation is the required standard form of a hyperbola with centre at (-1, -3)