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

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

#### Solution

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 xy = 0.

Concept: Tangents and Normals
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