Convert the polar equation to rectangular form.
r=4cos-4sin
What I have so far:
r^2=4rcos-4rsin
x^2+y^2-4x+4y
x^2-4x+y^2+4y=0
so, think back to your Algebra II days and finish it up. What you have is correct, but make it look good:
x^2-4x + y^2+4y = 0
x^2-4x+4 + y^2+4y+4 = 8
(x-2)^2 + (y+2)^2 = 8
http://www.wolframalpha.com/input/?i=r%3D4cos%CE%B8+-+4sin%CE%B8
To convert the given polar equation to rectangular form, we'll use the following relationships:
x = r * cos(theta)
y = r * sin(theta)
Given equation: r = 4 * cos(theta) - 4 * sin(theta)
Substituting x and y into the equation:
x^2 + y^2 - 4x + 4y = 0
Now, let's complete the square to write the equation in standard form:
(x^2 - 4x) + (y^2 + 4y) = 0
To complete the square for the x terms, we'll need to add (4/2)^2 = 4 to both sides:
(x^2 - 4x + 4) + (y^2 + 4y) = 4
Similarly, to complete the square for the y terms, we'll add (4/2)^2 = 4 to both sides:
(x^2 - 4x + 4) + (y^2 + 4y + 4) = 4 + 4
Rearranging the terms and simplifying:
(x - 2)^2 + (y + 2)^2 = 8
Thus, the rectangular form of the given polar equation r = 4 * cos(theta) - 4 * sin(theta) is:
(x - 2)^2 + (y + 2)^2 = 8.
To convert a polar equation to rectangular form, we need to express the equation in terms of x and y.
Let's continue from where you left off.
You correctly obtained the equation x^2 - 4x + y^2 + 4y = 0 from r^2 = 4rcosθ - 4rsinθ.
Now, let's complete the square to rewrite this equation in standard form:
Starting with the equation x^2 - 4x + y^2 + 4y = 0, we can group the x-terms and the y-terms separately:
(x^2 - 4x) + (y^2 + 4y) = 0
Now, let's complete the square for each group of terms individually:
(x^2 - 4x + 4) + (y^2 + 4y + 4) = 4 + 4
Simplifying further, we have:
(x - 2)^2 + (y + 2)^2 = 8
So, the equation in rectangular form is (x - 2)^2 + (y + 2)^2 = 8.