Question

Show that the normal at any point $\theta$ to the curve $x=acos\theta +a\theta ,y=asin\theta -a\theta cos\theta$ is. at a constant distance from the origin.

Answer 1

The given curve is $x=acos\theta +a\theta sin\theta ,y=asin\theta -a\theta cos\theta$ on differentiating w.r.t. $\theta$ we get
$\frac{dx}{d\theta }=-asin\theta +a[\theta cos\theta +sin\theta ]=a\theta cos\theta$
and $\frac{dy}{d\theta }=acos\theta -a\left[\theta \right(-sin\theta \left)\right]+cos\theta ]=acos\theta +a\theta sin\theta -acos\theta =a\theta sin\theta$
$\therefore$ 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
$y-[asin\theta -a\theta cos\theta ]=-cot\theta [x-(acos\theta +a\theta sin\theta \left)\right]\Rightarrow y-[asin\theta -a\theta cos\theta ]=-\frac{cos\theta }{sin\theta }[x-(acos\theta +a\theta sin\theta \left)\right]\Rightarrow ysin\theta -asi{n}^2\theta +a\theta sin\theta cos\theta =-xcos\theta +aco{s}^2\theta +a\theta sin\theta cos\theta \Rightarrow xcos\theta +ysin\theta =a(si{n}^2\theta +co{s}^2\theta )\Rightarrow xcos\theta +ysin\theta =a\Rightarrow xcos\theta +ysin\theta -a=0$
Now, the perpendicular distance of the normal from the origin is = $\frac{|-a|}{\sqrt{co{s}^2\theta +si{n}^2\theta }}=\frac{|-a|}{\sqrt{1}}=|-a|(\because co{s}^2\theta +si{n}^2\theta =1)$
Which is independant of $\theta .$ Hence, the perpendicular distance of the normal from the origin is constant