Question

Show that the equation of normal at any point on the curve $x=3\cos \theta -{\cos }^3\theta ,$ $y=3\sin \theta -{\sin }^3\theta$ is $4(y{\cos }^3\theta -x{\sin }^3\theta )=3\sin 4\theta$.

Answer 1

We have $x=3cos\theta -co{s}^3\theta$
therefore $\frac{dx}{d\theta }=-3sin\theta +3co{s}^2\theta sin\theta =-3sin\theta (1-co{s}^2\theta )=-3si{n}^3\theta$
$\frac{dy}{d\theta }=3cos\theta -3si{n}^2\theta cos\theta =3cos\theta (1-si{n}^2\theta )-3co{s}^2\theta$
$\frac{dy}{dx}=-\frac{co{s}^3\theta }{si{n}^3\theta }$ therefore , slope of normal $=+\frac{si{n}^3\theta }{co{s}^3\theta }$
hence the equation of normal is
$y-(3sin\theta -si{n}^3\theta )-\frac{si{n}^3\theta }{co{s}^3\theta }[x-(3cos\theta -co{s}^3\theta \left)\right]$
$\Rightarrow yco{s}^2\theta -3sin\theta co{s}^3\theta +si{n}^3\theta co{s}^3\theta =xsi{n}^3\theta -3si{n}^3\theta cos\theta +si{n}^3\theta co{s}^3\theta$

$\Rightarrow yco{s}^3\theta -xsi{n}^3\theta =3sin\theta cos\theta (co{s}^2\theta -si{n}^2\theta )$

$\Rightarrow yco{s}^3\theta -xsi{n}^3\theta =\frac{3}{2}sin2\theta .cos2\theta$

$\Rightarrow yco{s}^3\theta -xsi{n}^3\theta =\frac{3}{4}sin4\theta$
or $4(yco{s}^3\theta -xsi{n}^3\theta )=3sin4\theta$