The equation of a curve in polar coordinates | vedclass

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Question

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

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a parabola

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(B) Given the polar 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$$l = r(1 - \cos 	heta)$$l = r - r \cos 	heta$Since $x = r \cos 	heta$ and $r = \sqrt{x^2 + y^2}$,we substitute these into the equation:$l = \sqrt{x^2 + y^2} - x$$l + x = \sqrt{x^2 + y^2}$Squaring both sides:$(l + x)^2 = x^2 + y^2$$l^2 + 2lx + x^2 = x^2 + y^2$$y^2 = 2lx + l^2$$y^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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