How do I complete the square for this problem?? It's on translating conic sections.
x^2-4x+y^2+2y=-1
x^2-4x+y^2+2y=-1
x^2- 4x + 4 + y^2 + 2y + 1=-1 +4 + 1
(x-2)^2 + (y+1)^2 = 4
I assume you will recognize this as a circle.
x^=13
To complete the square for the given equation, we need to rearrange the terms and group the x-terms together and the y-terms together.
The given equation is:
x^2 - 4x + y^2 + 2y = -1
To complete the square for the x-terms, we will focus on the coefficient of x. In this case, it is -4.
To complete the square for any quadratic equation in the form x^2 + bx, we need to take half of the coefficient of x, square it, and add it to both sides of the equation.
So, for the x-terms, we will take half of -4, which is -2, square it (which is 4), and add 4 to both sides of the equation:
x^2 - 4x + 4 + y^2 + 2y = -1 + 4
Now, let's complete the square for the y-terms. We will focus on the coefficient of y. In this case, it is 2.
To complete the square for any quadratic equation in the form y^2 + dy, we need to take half of the coefficient of y, square it, and add it to both sides of the equation.
So, for the y-terms, we will take half of 2, which is 1, square it (which is 1), and add 1 to both sides of the equation:
x^2 - 4x + 4 + y^2 + 2y + 1 = -1 + 4 + 1
Now, let's simplify the equation:
x^2 - 4x + 4 + y^2 + 2y + 1 = 4
Combining like terms, we have:
(x^2 - 4x + 4) + (y^2 + 2y + 1) = 4
Now, notice that (x^2 - 4x + 4) can be factored as (x - 2)^2 and (y^2 + 2y + 1) can be factored as (y + 1)^2.
So, we have:
(x - 2)^2 + (y + 1)^2 = 4
This equation represents a circle with the center at (2, -1) and a radius of 2.