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: (C)

it makes angle $\pi/2 + \theta$ with the $x$-axis

Answer 2

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.