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

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Solution

At a point $t$ , the slope $\frac{- d x}{d y} = \frac{- d x}{d t} \times \frac{d t}{d y}$
$= - 1 \times (- 3 sin t + 3 {cos}^{2} t \times sin t) \times \frac{1}{(3 cos t - 3 {sin}^{2} t cos t)}$
$= sin t \times {sin}^{2} t \times \frac{1}{{cos}^{3} t}$
$= {tan}^{3} t$
Now, the equation of the line be $y = m x + c$ , where $m = {tan}^{3} t$
Since the line passes through the given point, we substitute the given values and get $3 sin t - {sin}^{3} t = 3 sin t \times {tan}^{2} t - {sin}^{3} t + c$
$\Rightarrow c = 3 sin t \times cos 2 t \times {sec}^{2} t$
The equation becomes $y = {tan}^{3} t x + 3 sin t \times cos 2 t \times {sec}^{2} t$
Multiplying throughout by $c o s^{3} t$ , we get $y {cos}^{3} t - x {sin}^{3} t = 3 sin t \times cos 2 t \times cos t$
i.e. $4 (y {cos}^{3} t - x {sin}^{3} t) = 3 sin 4 t$

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