If OT and ON are perpendicular 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

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The length of the normal is $|\frac{y}{\sin t}|$

The length of sub tangent is $|y \tan t|$

The length of the tangent is $|\frac{y}{\cos t}|$

$4 O T^{2} + O N^{2} = a^{2}$

answer is A, B, C, D.

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Detailed Solution

At any point t on the given curve, we have
$\frac{dy}{dx} = \frac{3 {acos}^{2} t (- \sin t)}{3 {asin}^{2} t (\cos t)} = - \cot t$
The equations of the tangent and normal at t are
$\begin{array}{l} xcot t + ysin t - (a / 2) \sin 2 t = 0 \\ And xsin t - ycos t - acos 2 t = 0 \\ OT = \frac{| (a / 2) \sin 2 t |}{\sqrt{\cos^{2} t + \sin^{2} t}} = \frac{a}{2} \sin 2 t \end{array}$
or 2OT=asin2t
$\begin{array}{l} and ON = \frac{| acos 2 t |}{\sqrt{\sin^{2} t + \cos^{2} t}} = acos 2 t \\ 4 {OT}^{2} + {ON}^{2} = a^{2} \sin^{2} 2 t + a^{2} \cos^{2} 2 t = a^{2} \end{array}$
Length of the tangent at $t = \frac{|y \sqrt{1 + {(\frac{dy}{dx})}^{2}}|}{| dy / dx |}$
$\begin{array}{l} = |\frac{ycosec t}{- \cot t}| \\ = |\frac{y}{\cos t}| \end{array}$
Length of the normal at $t = |y \sqrt{1 + {(\frac{dy}{dx})}^{2}}|$
$= |y \sqrt{1 + \cot^{2} t}| = |\frac{y}{\sin t}|$