Q99 / 225 · JEE Main 2005 · Mathematics

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

  • A
    it passes through the origin
  • B
    it makes angle $\frac{\pi}{2}+\theta$ with the $x$-axis
  • C
    it passes through $\left(\mathrm{a} \frac{\pi}{2},-\mathrm{a}\right)$
  • Correct answer
    D
    it is at a constant distance from the origin

Solution

Clearly $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta \Rightarrow$ slope of normal $=-\cot \theta$ Equation of normal at ' $\theta$ ' is $y-a(\sin \theta-\theta \cos \theta)=-\cot \theta(x-a(\cos \theta+\theta \sin \theta)$ $\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \cos \theta \sin \theta=-x \cos \theta+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$ $\Rightarrow x \cos \theta+y \sin \theta=a$ Clearly this is an equation of straight line which is at a constant distance 'a' from origin.

Hence, the correct answer is option: 'D'.

Relevant Questions