Find the Properties y^2-4y+4x+4=0

Problem

y2−4*y+4*x+4=0

Solution

1. Identify the type of conic section by observing that only one variable, y is squared, which indicates the equation represents a parabola.

2. Rearrange the equation to isolate the terms involving y on one side and the terms involving x and the constant on the other.

y2−4*y=−4*x−4

3. Complete the square for the y terms by adding ((−4)/2)2=4 to both sides of the equation.

y2−4*y+4=−4*x−4+4

4. Simplify both sides to write the equation in the standard form (y−k)2=4*p*(x−h)

(y−2)2=−4*x

5. Determine the vertex (h,k) from the standard form (y−2)2=−4*(x−0)

Vertex=(0,2)

6. Find the value of p by setting 4*p equal to the coefficient of the linear term.

4*p=−4

p=−1

7. Identify the direction of opening based on the squared variable and the sign of p Since y is squared and p<0 the parabola opens to the left.

8. Calculate the focus using the formula (h+p,k) for a horizontal parabola.

Focus=(0+(−1),2)=(−1,2)

9. Determine the directrix using the formula x=h−p

x=0−(−1)

x=1

10. Identify the axis of symmetry using the formula y=k

y=2

Final Answer

Properties: Vertex *(0,2), Focus *(−1,2), Directrix *x=1, Axis of Symmetry *y=2

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