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 ${}^{\prime }{\theta }^{\prime }$ 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
$y-asin\theta +a\theta cos\theta =-\frac{cos\theta }{sin\theta }\left(x-acos\theta -asin\theta \right)$
$\Rightarrow sin0y-asi{n}^2\theta +a\theta cos\theta sin\theta$
$=-xcos0+aco{s}^2\theta +a\theta sin\theta cos\theta$
$\Rightarrow xcos\theta +ysin\theta =a$
It is always at a constant distance ${}^{\prime }{a}^{\prime }$ from origin.