Question

Statement I The normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cdot \cos \theta$ is at a constant distance from the origin.
Statement II The perpendicular distance $(d)$ from origin to the straight line is
$d=\frac{\mid \text { constant } \mid}{\sqrt{(\text { Coefficient of } x )^2+(\text { Coefficient of } y )^2}}$

• A. Statement I is true; statement II is true; statement II is correct explanation of statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation of statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (A)

First, we find the equation of normal to the curve at any point $\theta$ and then find the perpendicular distance (d) from origin by using the relation
$d=\frac{\mid \text { constant } \mid}{\sqrt{(\text { coefficient of } x)^2+(\text { coefficient of } y)^2}}$
The given curve is $x=a \cos \theta+a \theta \sin \theta$, $y=a \sin \theta-a \theta \cos \theta$
On differentiating w.r.t. $\theta$, we get
$\frac{d x}{d \theta} =-a \sin \theta+a[\theta \cos \theta+\sin \theta] $
$ =-a \sin \theta+a \theta \cos \theta+a \sin \theta=a \theta \cos \theta $
and $ \frac{d y}{d \theta} =a \cos \theta-a[\theta(-\sin \theta)+\cos \theta] $
$=a \cos \theta +a \theta \sin \theta-a \cos \theta=a \theta \sin \theta $
$\therefore$ Slope of the tangent at $\theta$,
$\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$
Slope of the normal at $\theta=-\frac{1}{\frac{d y}{d x}}=-\frac{1}{\tan \theta}=-\cot \theta$
The equation of the normal at a given point $(x, y)$ is given by
$y-[a \sin \theta-a \theta \cos \theta]=-\cot \theta[x-(a \cos \theta+a \theta \sin \theta)]$
$\Rightarrow y-[a \sin \theta-a \theta \cos \theta]=-\frac{\cos \theta}{\sin \theta}$
$[x-(a \cos \theta+a \theta \sin \theta)]$
$\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \sin \theta \cos \theta=-x \cos \theta$
$+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$
$\Rightarrow x \cos \theta+y \sin \theta=a\left(\sin ^2 \theta+\cos ^2 \theta\right)$
$\Rightarrow x \cos \theta+y \sin \theta=a$
$\Rightarrow x \cos \theta+y \sin \theta-a=0$
Now, the perpendicular distance of the normal from the origin is
$ =\frac{|-a|}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}} $
$ =\frac{|-a|}{\sqrt{1}}=|-a| \left(\because \cos ^2 \theta+\sin ^2 \theta=1\right)$
which is independent of $\theta$. Hence, the perpendicular distance of the normal from the origin is constant.
$\therefore$ Both the statements are true and statement II is the correct explanation of statement I.