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
$\frac{dx}{d\theta }=-a\sin \theta +a[\theta \cos \theta +\sin \theta ]$
$=-a\sin \theta +a\theta \cos \theta +a\sin \theta =a\theta \cos \theta$
and $\frac{dy}{d\theta }=a\cos \theta -a[\theta (-\sin \theta )+\cos \theta ]$
$=a\cos \theta +a\theta \sin \theta -a\cos \theta =a\theta \sin \theta$
$\therefore$ Slope of the tangent at $\theta$,
$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta$
Slope of the normal at $\theta =-\frac{1}{\frac{dy}{dx}}=-\frac{1}{\tan \theta }=-\cot \theta$
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.