Question

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 of the following is/are correct?

• A. $4{OT}^2+ON{}^2=a^2$
• B. ${OT}^2+{ON}^2=a^2$
• C. ${OT}^2+{ON}^2=2a^2$
• D. ${OT}^2+2{ON}^2=4a^2$

Answer 1

The correct option is A $4{OT}^2+ON{}^2=a^2$

Given equation of curve is

$x=a{\sin }^{3}t,y=a{\cos }^{3}t$

$\frac{dx}{dt}=3a{\sin }^{2}t\cos t$

$\frac{dy}{dt}=-3a{\cos }^{2}t\sin t$

$\Rightarrow \frac{dy}{dx}=-\cot t$

Slope of tangent to curve at $t$ is $-\cot t$

Equation of tangent at $t$ is

$y-a{\cos }^{3}t=-\cot t(x-a{\sin }^{3}t)$

$\Rightarrow -x\cot t-y+a{\cos }^{3}t+a\cos t{\sin }^{2}t=0$

Now, $OT=\left|\frac{a{\cos }^{3}t+a\cos t{\sin }^{2}t}{\sqrt{1+{\cot }^{2}t}}\right|$

$\Rightarrow OT=|a\cos t{\sin }^{3}t+a{\cos }^{3}t\sin t|$

$\Rightarrow O{T}^2=a^2{\sin }^{2}t{\cos }^{2}t({\sin }^{2}t+{\cos }^{2}t{)}^2$

$\Rightarrow 4O{T}^2=4a^2{\sin }^{2}t{\cos }^{2}t$ ......(1)

Now, slope of normal to curve at $t$ is $\tan t$

Equation of normal at $t$ is

$y-a{\cos }^{3}t=\tan t(x-a{\sin }^{3}t)$

$\Rightarrow x\tan t-y+a{\cos }^{3}t-a\tan t{\sin }^{3}t=0$

Now, $ON=\left|\frac{a{\cos }^{3}t-a\tan t{\sin }^{3}t}{\sqrt{1+{\tan }^{2}t}}\right|$

$\Rightarrow ON=|a{\cos }^{4}t-a{\sin }^{4}t|$

$\Rightarrow O{N}^2=a^2({\cos }^{4}t-{\sin }^{4}t{)}^2$

$\Rightarrow O{N}^2=a^2({\cos }^{2}t-{\sin }^{2}t{)}^2$

$\Rightarrow O{N}^2=a^2({\cos }^{2}t+{\sin }^{2}t{)}^2-4a^2{\sin }^{2}t{\cos }^{2}t$

$\Rightarrow O{N}^2=a^2-4a^2{\sin }^{2}t{\cos }^{2}t$ .....(2)

Adding (1) and (2), we get

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

Hence, option A.