If the tangent and normal drawn to the curve | vedclass

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(C) Given curve is $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.First, we find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \

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$a^2$

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(C) Given curve is $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.First, we find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \sin 	heta}{a(1+\cos 	heta)} = 	an \frac{	heta}{2}$.At $P\left(	heta=\frac{\pi}{2}ight)$, the slope of the tangent $m_T = 	an \frac{\pi}{4} = 1$ and the slope of the normal $m_N = -1$.The coordinates of point $P$ are $x = a(\frac{\pi}{2} + 1)$ and $y = a(1 - 0) = a$.Equation of the tangent at $P$: $y - a = 1(x - a(\frac{\pi}{2} + 1))$. Setting $y=0$, we get $x = a(\frac{\pi}{2} + 1) - a = \frac{a\pi}{2}$. Thus, $A = (\frac{a\pi}{2}, 0)$.Equation of the normal at $P$: $y - a = -1(x - a(\frac{\pi}{2} + 1))$. Setting $y=0$, we get $-a = -x + a(\frac{\pi}{2} + 1)$, so $x = a(\frac{\pi}{2} + 2)$. Thus, $B = (a(\frac{\pi}{2} + 2), 0)$.The area of $	riangle PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |x_B - x_A| 	imes y_P$.Area $= \frac{1}{2} 	imes |a(\frac{\pi}{2} + 2) - \frac{a\pi}{2}| 	imes a = \frac{1}{2} 	imes |2a| 	imes a = a^2$.

If the tangent and normal drawn to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $P\left(\theta=\frac{\pi}{2}\right)$ cut the $X$-axis at $A$ and $B$ respectively, then the area (in sq. units) of $\triangle P A B$ is

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