The equation of normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is

The equation of normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is
  1. $2 x+2 y-\pi=0$
  2. $2 x-y-\pi=0$
  3. $2 x-2 y-\pi=0$
  4. $2 x+y-\pi=0$

Solution

$\begin{array}{ll} & x=\theta+\sin \theta \text { and } y=1+\cos \theta \\ \therefore \quad & \frac{\mathrm{d} x}{\mathrm{~d} \theta}=1+\cos \theta \text { and } \frac{\mathrm{d} y}{\mathrm{~d} \theta}=-\sin \theta \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{~d} x}{\mathrm{~d}} \theta}=\frac{-\sin \theta}{1+\cos \theta} \\ & \text { At } \theta=\frac{\pi}{2} \\ & x=\frac{\pi}{2}+\sin \frac{\pi}{2}=\frac{\pi}{2}+1 \text { and } \\ & y=1+\cos \frac{\pi}{2}=1\end{array}$ $\therefore \quad\left(\frac{d y}{d x}\right)_{\left(\theta=\frac{\pi}{2}\right)}=\frac{-\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=-1$ $\therefore \quad$ Slope of normal $=1$ $\therefore \quad$ Equation of the normal at $\left(\frac{\pi}{2}+1,1\right)$ is $\begin{aligned} & y-1=1\left(x-\frac{\pi}{2}-1\right) \\ & \Rightarrow 2 y-2=2 x-\pi-2 \\ & \Rightarrow 2 x-2 y-\pi=0 \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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