Question

The normal of the curve $x=a(cos\theta +\theta sin\theta );y=a(sin\theta -\theta cos\theta )$ at any $\theta$ is such that

• A. It makes a constant angle with x-axis
• B. It passes through the origin
• C. It is at a constant distance from the origin
• D. None of these

Answer 1

The correct option is C It is at a constant distance from the origin

$y=a(sin\theta -\theta cos\theta ),x=a(cos\theta +\theta sin\theta )$
$\frac{dy}{d\theta }=a[cos\theta -cos\theta +\theta sin\theta ]=a\theta sin\theta \frac{dx}{d\theta }=a(-sin\theta +sin\theta +\theta cos\theta )=a\theta cos\theta \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\theta sin\theta }{a\theta cos\theta }=tan\theta$
$\Rightarrow$ Slope of the tangent = $tan\theta$
$\therefore$ Slope of the normal =$-cot\theta$
Hence, equation of normal
$[y-asin\theta +a\theta cos\theta ]=-\frac{cos\theta }{sin\theta }[x-acos\theta -a\theta sin\theta ]\Rightarrow ysin\theta -asi{n}^2\theta +a\theta sin\theta cos\theta =-xcos\theta +aco{s}^2\theta +a\theta sin\theta cos\theta \Rightarrow xcos\theta +ysin\theta =a(si{n}^2\theta +co{s}^2\theta )\Rightarrow xcos\theta +ysin\theta =a$
$\therefore$ Distance from origin =$\frac{a}{\sqrt{si{n}^2\theta +co{s}^2\theta }}=a$= constant