Question

Find the equations of the tangent and normal to the curve x = a sin³θ and y = a cos³θ at θ=π/4.

Answer 1

Given: ` x=a sin^3θ and y=a cos^3θ`

`⇒dx/dθ=3a sin^2θ cosθ and dy/dθ=−3a cos^2θ sinθ`

`⇒dy/dx=(dy/dθ)/(dx/dθ)`

`=(−3a cos^2θ sinθ)/(3a sin^2θ cosθ)`

`=-costheta/sintheta`

So, the equation of the tangent is given by

`y−a cos^3θ=−cosθ/sinθ(x−a sin^3θ)  [Using  y−y1=(dy/dx)_(x1,y1)(x−x1)]`

Putting θ=π/4, we get:

`y-a(1/sqrt2)^3=-(1/sqrt2)/(1/sqrt2)(x-a(1/sqrt2)^3)     (∵ sin(pi/4)=1/sqrt2, cos(pi/4)=1/sqrt2`

 

`y-a1/(2sqrt2)=-1xx(x-a/(2sqrt2))`

`x+y=a/(2sqrt2)+a/(2sqrt2)=a/sqrt2`

`x+y=a/sqrt2 or sqrt2x+sqrt2y=a`

Thus, the equation of the tangent at `θ=pi/4 is sqrt2x+sqrt2y=a.`

Now, the equation of the normal is given by

`y-acos^3theta=-1/(-costheta/sintheta)(x-asin^3theta) [`

Putting θ=π/4, we get:

`y-a(1/sqrt2)^3=-1/((-1/sqrt2)/(1/sqrt2))(x-a(1/sqrt2)^3)  (∵ sin(pi/4)=1/sqrt2, cos(pi/4)=1/sqrt2)`

 

`y-a(1/2sqrt2)=1xx(x-a/(2sqrt2))`

`=x-y=a/(2sqrt2)-a/(2sqrt2)=0`

x-y=0

∴ The equation of the normal at θ=π/4 is x − y = 0.