Application of Derivatives 1 Question 13
####13. 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
(1983, 1M)
(a) it makes a constant angle with the $X$-axis
(b) it passes through the origin
(c) it is at a constant distance from the origin
(d) None of the above
Objective Questions II
(One or more than one correct option)
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Answer:
Correct Answer: 13. $x+2 y=\frac{\pi}{2}$ and $x+2 y=\frac{-3 \pi}{2}$
Solution:
- Given, $x=a(\cos \theta+\theta \sin \theta)$
and $y=a(\sin \theta-\theta \cos \theta)$
$\therefore \quad \frac{d x}{d \theta}=a(-\sin \theta+\sin \theta+\theta \cos \theta)=a \theta \cos \theta$
and $\frac{d y}{d \theta}=a(\cos \theta-\cos \theta+\theta \sin \theta)$
$ \frac{d y}{d \theta}=a \theta \sin \theta \Rightarrow \frac{d y}{d x}=\tan \theta $
Thus, equation of normal is
$ \frac{y-a(\sin \theta-\theta \cos \theta)}{x-a(\cos \theta+\theta \sin \theta)}=\frac{-\cos \theta}{\sin \theta} $
$\Rightarrow-x \cos \theta+a \theta \sin \theta \cos \theta+a \cos ^{2} \theta$ $=y \sin \theta+\theta a \sin \theta \cos \theta-a \sin ^{2} \theta$
$\Rightarrow \quad x \cos \theta+y \sin \theta=a$
whose distance from origin is
$ \frac{|0+0-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=a $
