Question

How do I go about converting the general form 4x^2 + 6y = -4y^2 + 12 to its standard form?

Answer 1

Step-by-step explanation:

Given:

`4x^2+6y=-4y^2+12`

To find:

convert from general to standard form

The standard form of a circle:

`(x\text{ - a\rparen}^2\text{ + \lparen y - b\rparen}^2\text{ = r}^2`

To convert the given form to the standard form, we will write x and y into perfect squares using the complete the square:

`\begin{gathered} first\text{ divide through by 4 \lparen we need the coefficient of x}^2\text{ and y}^2\text{ to be 1\rparen} \\ (4x^2)/(4)\text{ + }(6y)/(4)\text{ = }(-4y^2)/(4)\text{ + }(12)/(4) \\ \\ x^2\text{ + }(3)/(2)y\text{ = -y}^2\text{ + 3} \\ \\ x^2\text{ + y}^2\text{ +}(3)/(2)y\text{ = 3} \end{gathered}`
`\begin{gathered} for\text{ x, we only have x}^2.\text{ There is no coefficient of x. We can't complete the square} \\ \\ for\text{ y: }y^2\text{ + }(3)/(2)y \\ We\text{ will complete the square:} \\ coefficient\text{ of y = 3/2; half the coefficient of y = 3/4} \\ square\text{ of half the coefficient of y = \lparen}(3)/(4))\placeholder{⬚}^2 \\ \\ Add\text{ square of half the coefficient of y to both sides of the equation:} \\ x^2\text{ + y}^2\text{ + }(3)/(2)y\text{ + \lparen}(3)/(4))\placeholder{⬚}^2\text{ = 3 + \lparen}(3)/(4))\placeholder{⬚}^2 \\ \\ x^2\text{ + \lparen y + }(3)/(4))\placeholder{⬚}^2\text{ = 3 + }(9)/(16) \end{gathered}`
`\begin{gathered} x^2\text{ + \lparen y + }(3)/(4))\placeholder{⬚}^2\text{ = }\frac{3(16)\text{ + 9}}{16} \\ \\ x^2\text{ + \lparen y + }(3)/(4))\placeholder{⬚}^2\text{ = }(57)/(16) \end{gathered}`