standard form of 16x^2 + 4y^2 + 24y - 28 = 0?
Answer (1)
Standard formStep 1: Group like terms and move the constant to the right side.[tex]$\sf{16x^2 + 4y^2 + 24y - 28 = 0}$[/tex]Group the y-terms and move the constant to the right[tex]$\sf{16x^2 + (4y^2 + 24y) = 28}$[/tex]Step 2: Complete the square for the y-terms.Factor out the coefficient of the [tex]$\sf{y^2}$[/tex] term (which is 4) from the y-terms[tex]$\sf{16x^2 + 4(y^2 + 6y) = 28}$[/tex]To complete the square for [tex]$\sf{y^2 + 6y}$[/tex], we need to add and subtract [tex]$\sf{( \frac{6}{2})^2 = 3^2 = 9}$[/tex] inside the parenthesis[tex]$\sf{16x^2 + 4(y^2 + 6y + 9 - 9) = 28}$[/tex]16x^2 + 4((y + 3)^2 - 9) = 28[tex]$\sf{16x^2 + 4(y + 3)^2 - 36 = 28}$[/tex]Step 3: Isolate the squared terms and constant.Move the constant term to the right side:[tex]$\sf{16x^2 + 4(y + 3)^2 = 28 + 36}$[/tex][tex]$\sf{16x^2 + 4(y + 3)^2 = 64}$[/tex]Step 4: Divide to get 1 on the right side.Divide both sides of the equation by 64:[tex]$\sf{\frac{(16x^2)}{64} + \frac{(4(y + 3)^2)}{64} = \frac{64}{64} }$[/tex][tex]$\sf{\frac{x^2}{4} + \frac{(y + 3)^2}{16} = 1}$[/tex]Final answer:The standard form of the given equation is [tex]$\sf{\frac{x^2}{4} + \frac{(y + 3)^2}{16} = 1}$[/tex]
