Graph x^2+4x+4y^2-8y+4=0

Problem

x^2+4*x+4*y^2−8*y+4=0

Solution

1. Group the x terms and y terms together and move the constant to the other side of the equation.

(x^2+4*x)+(4*y^2−8*y)=−4

2. Factor out the coefficient of the y^2 term to prepare for completing the square.

(x^2+4*x)+4*(y^2−2*y)=−4

3. Complete the square for both the x and y expressions by adding (b/2)^2 to both sides. For x add (4/2)^2=4 For y add ((−2)/2)^2=1 inside the parentheses, which is $4(1) = 4$ on the right side.

(x^2+4*x+4)+4*(y^2−2*y+1)=−4+4+4

4. Simplify the equation into its squared binomial forms.

(x+2)^2+4*(y−1)^2=4

5. Divide both sides by 4 to put the equation into the standard form of an ellipse, ((x−h)^2)/(a^2)+((y−k)^2)/(b^2)=1

((x+2)^2)/4+((y−1)^2)/1=1

6. Identify the properties of the ellipse. The center (h,k) is (−2,1) The horizontal semi-axis is a=√(,4)=2 and the vertical semi-axis is b=√(,1)=1

Final Answer

((x+2)^2)/4+((y−1)^2)/1=1

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