Which is a polar form of the following parametric equations ?

Question

x=4sin^2 theta
Y=4 sin theta cos theta

a) r=1/4sin theta cod theta
b) r=2
c)r=16sin^2 theta
d)r=4sin theta

Answer 1

To find the polar form of the given parametric equations, we need to convert the Cartesian equations into polar form.

Given:
x = 4sin^2(theta)
y = 4sin(theta)cos(theta)

From polar coordinates to Cartesian coordinates, we know:
x = r*cos(theta)
y = r*sin(theta)

Substitute the given equations into these formulas:
r*cos(theta) = 4sin^2(theta)
r*sin(theta) = 4sin(theta)cos(theta)

Divide the two equations to eliminate r:
r*cos(theta)/(r*sin(theta)) = 4sin^2(theta)/(4sin(theta)cos(theta))
tan(theta) = tan(theta)
This is true for all values of theta.

Therefore, the polar form of the parametric equations is:
r = 4sin(theta)

So, the correct answer is:
d) r = 4sin(theta)