Question

The normal to the curve $x=a\left(cos\,\theta+\theta\,sin\,\theta \right). y=a\left(sin\,\theta -\theta\,cos\,\theta \right)$ at any point $'\theta'$ is such that :

• A. it is at a constant distance from the origin
• B. it passes through $(a\, \pi/2, - a)$
• C. it makes angle $\pi/2 + \theta$ with the $x$-axis
• D. it passes through the origin

Answer 1

Answer: (A), (C)

Key Idea : Equation of normal at any point $(x_1 y_1)$ on any curve is
$y-y_{1}=-\frac{1}{\left(\frac{dy}{dx}\right)_{_{\left(x_1, y_1\right)}}} \left(x-x_{1}\right)$
Given that $x = a \left(cos\,\theta + \theta\,sin\,\theta\right)$
$\frac{dx}{d\theta}=a\left(-sin\,\theta+sin\,\theta +\theta\,cos\,\theta \right)$
$\Rightarrow \frac{dx}{d\theta }=a\,\theta \,cos\,\theta$
and $y=a\left(sin\,\theta -\theta\,cos\,\theta \right)$
$\Rightarrow \frac{dy}{d\theta}=a\,\theta \,sin\,\theta$
$\therefore \frac{dx}{dy}=-\frac{dy/d\theta}{dx/d\theta }=tan\,\theta $
Slope of normal
$=-\frac{dx}{dy}=-cot\,\theta=tan\left(\frac{\pi}{2}+\theta\right)$
So equation of normal is
$y-a\,sin\,\theta+a\,\theta\,cos\,\theta= -\frac{cos\theta}{sin\,\theta}\left(x-a\,cos\,\theta-a\,sin\,\theta\right)$
$\Rightarrow sin\,0y-a\,sin^{2}\,\theta+a\,\theta\,cos\,\theta\,sin\,\theta$
$=-x\,cos\,0+a\,cos^{2}\,\theta+a\,\theta\,sin\,\theta cos\,\theta$
$\Rightarrow x\,cos\,\theta+y\,sin\,\theta=a$
It is always at a constant distance $'a'$ from origin.