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

Problem

y2−4*x+4*y−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 opening horizontally.

2. Rearrange the equation to isolate the terms involving y on one side and the terms involving x and constants 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. Factor the left side as a perfect square and simplify the right side.

(y+2)2=4*x+8

5. Factor out the coefficient of x on the right side to put the equation into the standard form (y−k)2=4*p*(x−h)

(y+2)2=4*(x+2)

6. Determine the vertex (h,k) from the standard form, which is (−2,−2)

7. Calculate the value of p by setting 4*p=4 which gives p=1 Since p>0 the parabola opens to the right.

8. Find the focus and directrix using p The focus is (h+p,k)=(−1,−2) and the directrix is x=h−p=−3

Final Answer

(y+2)2=4*(x+2)

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