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

Problem

4*x2+y2−8*x+4*y+4=0

Solution

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

(4*x2−8*x)+(y2+4*y)=−4

2. Factor out the coefficient of the squared terms where necessary to prepare for completing the square.

4*(x2−2*x)+(y2+4*y)=−4

3. Complete the square for both the x and y expressions by adding (b/2)2 inside the parentheses and adding the equivalent values to the right side.

4*(x2−2*x+1)+(y2+4*y+4)=−4+4*(1)+4

4. Simplify the equation by writing the quadratic expressions as perfect squares and combining the constants on the right.

4*(x−1)2+(y+2)2=4

5. Divide the entire equation by 4 to put it into the standard form of an ellipse equation, ((x−h)2)/(a2)+((y−k)2)/(b2)=1

((x−1)2)/1+((y+2)2)/4=1

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

Final Answer

((x−1)2)/1+((y+2)2)/4=1

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