Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B it makes an angle $\frac{\pi }{2}+\theta$ with the x-axis
C it is at a constant distance from the origin

$\frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\tan \theta =$Slope of tangent

$\therefore$ Slope of normal to the curve$=-\cot \theta =\tan (90+\theta )$

Now, equation of normal to the curve $[y-a(\sin \theta -\theta \cos \theta )]=\frac{-\cos \theta }{\sin \theta }(x-a(\cos \theta +a\sin \theta ))$

$\Rightarrow x\cos \theta +y\sin \theta =a\left(1\right)$

Now, distance from $(0,0)$ to $x\cos \theta +y\sin \theta =a$ is

distance $=d=\frac{0+0-a}{1}=a$

$\therefore$ Distance is constant$=a$