convert to polar form:
1) x^2-y^2=4 (You may leave your answer for this one as r2 =)
2) x^2+(y-1)^2=1
we know
r^2 = x^2 + y^2
x/r = cosØ, and y/r = sinØ or x=rcosØ and y = rsinØ
x^2 - y^2 = 4
r^2 cos^2 Ø - r^2 sin^2Ø = 4
r^2 (cos^2 Ø - sin^2 Ø) = 4
r^2 (cos 2Ø)=4
r^2 = 4/cos 2Ø or r^2 = 4sec 2Ø
follow the same procedure for the second one, let me know what you got
how did you get cos 2Ø after subtracting cos^2Ø-sin^2Ø?
One of the basic trig identities says:
cos 2A = cos^2 A - sin^2 A
To convert equations to polar form, we need to express them in terms of polar coordinates, (r, θ). These coordinates represent a point in the polar coordinate system, where r represents the distance from the origin and θ represents the angle made with the positive x-axis.
Let's try to convert the given equations to polar form:
1) x^2 - y^2 = 4
To convert this equation to polar form, we can use the relationships between Cartesian and polar coordinates. In Cartesian coordinates, x = r * cos(θ) and y = r * sin(θ).
Substituting these values into the equation, we get:
(r * cos(θ))^2 - (r * sin(θ))^2 = 4
Simplifying this equation further, we have:
r^2 * cos^2(θ) - r^2 * sin^2(θ) = 4
Factoring out r^2, we get:
r^2 * (cos^2(θ) - sin^2(θ)) = 4
Using the identity cos^2(θ) - sin^2(θ) = cos(2θ), we can simplify the equation to:
r^2 * cos(2θ) = 4
Therefore, the equation in polar form becomes:
r^2 = 4/cos(2θ)
2) x^2 + (y - 1)^2 = 1
Using the same approach as above, we substitute x = r * cos(θ) and y = r * sin(θ) into the equation:
(r * cos(θ))^2 + (r * sin(θ) - 1)^2 = 1
Simplifying and expanding the equation, we have:
r^2 * cos^2(θ) + r^2 * sin^2(θ) - 2r * sin(θ) + 1 = 1
Since cos^2(θ) + sin^2(θ) = 1, the equation becomes:
r^2 = 2r * sin(θ)
Therefore, the equation in polar form is:
r = 2sin(θ)
Keep in mind that these are the polar forms of the given equations. If you need the solutions for specific values of r and θ, further calculations may be required.