If $OT$ and $ON$ are perpendiculars dropped from the origin to the tangent and normal to the curve $x = a {\sin }^{3}t, y = a {\cos }^{3}t$ at an arbitrary point, then which one of the following is correct?

At any point $\text{'}t\text{'}$ on the given curve, we have

     $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{3a{\cos }^{2}t(-\sin t)}{3a{\sin }^{2}t(\cos t)}=-\cot t$

The equations of the tangent and normal at $\text{'}t\text{'}$ are

      $x\cos t+y\sin t-\frac{a}{2}\sin 2t=0$                  (i)

and,  $x\sin t-y\cos t+a\cos 2t=0$               (ii)

$\therefore OT=\frac{\left|\right(a/2)\sin 2t|}{\sqrt{{\cos }^{2}t+{\sin }^{2}t}}=\frac{a}{2}\sin 2t\Rightarrow 2OT=a\sin 2t$

and, $ON=\frac{|a\cos 2t|}{\sqrt{{\sin }^{2}t+{\cos }^{2}t}}=a\cos 2t$

Hence,

    $4O{T}^2+O{N}^2=a^2{\sin }^{2}2t+a^2{\cos }^{2}2t=a^2$

Length of the tangent at $\text{'}t\text{'}$ $=\frac{\left|y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right|}{dy/dx}$

$\Rightarrow$  Length of the tangent at $\text{'}t\text{'}$$=\left|\frac{y\mathrm{cosec}t}{-\cot t}\right|=\left(\frac{y}{\cos t}\right)$

       Length of the normal at $\text{'}t\text{'}=\left|y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right|$

$\Rightarrow$ Length of the normal at $\text{'}t\text{'}=\left|y\sqrt{1+{\cot }^{2}t}\right|=\left|-\frac{y}{\mid \sin t}\right|$

Hence, options $\left(a\right), \left(b\right)$ and $\left(c\right)$ are correct.