The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$, where the…
The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$, where the tangent is inclined at an angle $\frac{\pi}{4}$ to $x$-axis, are
$\left[a\left(\frac{\pi}{4}-1\right), a\right]$
$\left[a\left(\frac{\pi}{2}+1\right), a\right]$
$\left(a \frac{\pi}{2}, a\right)$
$(a, a)$
Solution
Given coordinate is
$x=a(\theta+\sin \theta), y=a(1-\cos \theta)$
On differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d x}{d \theta} & =a(1+\cos \theta), \frac{d y}{d \theta}=a(0+\sin \theta) \\ \therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{a \sin \theta}{a(1+\cos \theta)} \\ & =\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}=\tan \frac{\theta}{2}\end{aligned}$
$\begin{aligned} & \tan \frac{\pi}{4}=\tan \frac{\theta}{2} \quad\left(\because \frac{d y}{d x}=\tan x\right) \\ & \Rightarrow \quad \frac{\pi}{4}=\frac{\theta}{2} \\ & \Rightarrow \quad \theta=\frac{\pi}{2} \\ & \end{aligned}$
$\therefore$ Coordinate of
$\begin{aligned} & P\left[a\left(\frac{\pi}{2}+\sin \frac{\pi}{2}\right), a\left(1-\cos \frac{\pi}{2}\right)\right] \\ = & P\left[a\left(\frac{\pi}{2}+1\right), a\right]\end{aligned}$