Question

Find the equation of tangent and normal at t, to the curve $x=asi{n}^{3}t,y=aco{s}^{3}t$

Answer 1

Curve, $x=asi{n}^{3}t................\left(1\right)$

$y=aco{s}^{3}t...............\left(2\right)$

Now, $\frac{dy}{dt}=a\left(3{cos}^{2}t\right)(-sint)$

and, $\frac{dx}{dt}=a\left({3sin}^{2}t\right)\left(cost\right)$

$=>\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-{cos}^{2}t}{{sin}^{2}t}.\frac{sint}{cost}=-cott.$

So, equation of tangent at $t$, $(y-aco{s}^{3}t)=\frac{dy}{dx}(x-asi{n}^{3}t)$

$=>y-a{cos}^{3}t=(-cott)x-a{sin}^{2}tcost$

$=>y+cottx+acost-2aco{s}^{3}t=0$

And normal at $t,=>(y-aco{s}^{3}t)=\frac{-1}{\frac{dy}{dx}}=(x-asi{n}^{3}t)$

$=>y-\left(tant\right)x-aco{s}^{3}t-asi{n}^{3}ttant=0$