How does one reduce the conic $4x^2+16x+y^2-6y+21=0$ to standard form?
How does one reduce the conic $4x^2+16x+y^2-6y+21=0$ to standard form?
Complete the square: $4x^2+ 16x= 4(x^2+ 4x)= 4(x^2+ 4x+ 4- 4)= 4(x+ 2)^2- 16$ and $y^2- 6y= y^2- 6y+ 9- 9= (y- 3)^2- 9$.