Find the Vertex Form 4x^2+24x+16y^2-32y-12=0

Problem

4*x2+24*x+16*y2−32*y−12=0

Solution

1. Group the terms containing x and the terms containing y and move the constant to the right side of the equation.

4*x2+24*x+16*y2−32*y=12

2. Factor out the leading coefficients of the x2 and y2 terms to prepare for completing the square.

4*(x2+6*x)+16*(y2−2*y)=12

3. Complete the square for both the x and y expressions. For x add (6/2)2=9 inside the parentheses. For y add ((−2)/2)2=1 inside the parentheses. Balance the equation by adding 4*(9) and 16*(1) to the right side.

4*(x2+6*x+9)+16*(y2−2*y+1)=12+36+16

4. Simplify the right side and write the quadratic expressions as perfect squares.

4*(x+3)2+16*(y−1)2=64

5. Divide the entire equation by 64 to set the right side to 1 which is the standard form for an ellipse (the "vertex form" for this conic section).

(4*(x+3)2)/64+(16*(y−1)2)/64=1

6. Reduce the fractions to reach the final form.

((x+3)2)/16+((y−1)2)/4=1

Final Answer

4*x2+24*x+16*y2−32*y−12=0⇒((x+3)2)/16+((y−1)2)/4=1

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