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