Find the Vertex Form x^2-y^2-2x-4y-4=0

Problem

x^2−y^2−2*x−4*y−4=0

Solution

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

(x^2−2*x)−(y^2+4*y)=4

2. Complete the square for the x terms by adding ((−2)/2)^2=1 inside the first set of parentheses.

(x^2−2*x+1)−(y^2+4*y)=4+1

3. Complete the square for the y terms by adding (4/2)^2=4 inside the second set of parentheses. Note that because of the leading negative sign, adding 4 inside the parentheses subtracts 4 from the left side, so subtract 4 from the right side as well.

(x^2−2*x+1)−(y^2+4*y+4)=4+1−4

4. Factor the perfect square trinomials into binomial squares.

(x−1)^2−(y+2)^2=1

5. Isolate the y term to express the equation in the standard vertex form for a hyperbola or a conic section.

(y+2)^2=(x−1)^2−1

Final Answer

x^2−y^2−2*x−4*y−4=0⇒(x−1)^2−(y+2)^2=1

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