Convert cartesian equation to polar form
1. x^2+y^2=4x
2. (x^2+y^2)^2=6(x^2-y^2)
These take some practice, but you soon learn the tricks of thinking...
#1
r^2 = 4r cosθ
r = 4cosθ
#2
(r^2)^2 = 6(r^2 cos^2θ - r^2 sin^2θ)
r^2 = 6(cos^2θ - sin^2θ)
r^2 = 6cos(2θ)
To convert the given equations from Cartesian form to polar form, we need to express the variables (x and y) in terms of polar variables (r and θ).
1. x^2 + y^2 = 4x:
First, we can rewrite this equation as:
x^2 - 4x + y^2 = 0
Now, we can express x and y in terms of polar variables:
x = r * cos(θ)
y = r * sin(θ)
Substituting these expressions into the equation, we have:
(r * cos(θ))^2 - 4(r * cos(θ)) + (r * sin(θ))^2 = 0
Simplifying, we get:
r^2 * cos^2(θ) - 4r * cos(θ) + r^2 * sin^2(θ) = 0
Using the trigonometric identity: cos^2(θ) + sin^2(θ) = 1, we can simplify further:
r^2 - 4r * cos(θ) = 0
Dividing both sides by r, we have:
r - 4 * cos(θ) = 0
Therefore, the polar form of the equation x^2 + y^2 = 4x is:
r = 4 * cos(θ)
2. (x^2 + y^2)^2 = 6(x^2 - y^2):
Similarly, we can rewrite this equation as:
x^4 + 2x^2y^2 + y^4 = 6x^2 - 6y^2
Again, converting x and y into polar variables:
x = r * cos(θ)
y = r * sin(θ)
Substituting these expressions into the equation, we get:
(r * cos(θ))^4 + 2(r * cos(θ))^2 * (r * sin(θ))^2 + (r * sin(θ))^4 = 6(r * cos(θ))^2 - 6(r * sin(θ))^2
Simplifying, we have:
r^4 * cos^4(θ) + 2r^2 * cos^2(θ) * sin^2(θ) + r^4 * sin^4(θ) = 6r^2 * cos^2(θ) - 6r^2 * sin^2(θ)
Using the trigonometric identity: cos^2(θ) + sin^2(θ) = 1, we can simplify further:
r^4 * (cos^4(θ) + sin^4(θ)) + 2r^2 * cos^2(θ) * sin^2(θ) = 6r^2 * (cos^2(θ) - sin^2(θ))
Again, using the same identity cos^2(θ) + sin^2(θ) = 1, we have:
r^4 + 2r^2 * cos^2(θ) * sin^2(θ) = 6r^2 * (cos^2(θ) - sin^2(θ))
Dividing both sides by r^2, we get:
r^2 + 2 * cos^2(θ) * sin^2(θ) = 6 * (cos^2(θ) - sin^2(θ))
Finally, simplifying the equation, we have:
r^2 - 4 * cos^2(θ) * sin^2(θ) = 0
Therefore, the polar form of the equation (x^2 + y^2)^2 = 6(x^2 - y^2) is:
r^2 = 4 * cos^2(θ) * sin^2(θ)