The equation of a curve in polar coordinates | vedclass

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Question

(B) Given the equation $\frac{l}{r} = 2 \sin^2 \frac{	heta}{2}$.Using the trigonometric identity $2 \sin^2 \frac{	heta}{2} = 1 - \cos 	heta$,we get

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a parabola

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(B) Given the equation $\frac{l}{r} = 2 \sin^2 \frac{	heta}{2}$.Using the trigonometric identity $2 \sin^2 \frac{	heta}{2} = 1 - \cos 	heta$,we get $\frac{l}{r} = 1 - \cos 	heta$.Rearranging,$l = r(1 - \cos 	heta) = r - r \cos 	heta$.Since $x = r \cos 	heta$ and $r = \sqrt{x^2 + y^2}$,we have $l = \sqrt{x^2 + y^2} - x$.Thus,$\sqrt{x^2 + y^2} = x + l$.Squaring both sides,$x^2 + y^2 = (x + l)^2 = x^2 + 2lx + l^2$.Simplifying,$y^2 = 2lx + l^2 = 2l(x + \frac{l}{2})$.This is the standard form of a parabola $Y^2 = 4AX$ with vertex at $(-\frac{l}{2}, 0)$.

The equation of a curve in polar coordinates is $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$. This equation represents:

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