Given as x = θ + sinθ, y = 1 + cosθ at θ = π/2
Differentiate the given equation with respect to θ, to get the slope of the tangent
dx/dθ = 1 + cosθ
dy/dθ = -sinθ
On dividing both the above equation
dy/dx = - sinθ/(1 + cosθ)
m(tangent) at θ = (π/2) = - 1
The normal is perpendicular to tangent therefore, m1m2 = – 1
m(normal) at θ = (π/2) = 1
The equation of tangent is given by y – y1 = m(tangent)(x – x1)
y - 1 = -1(x - (π/2) -1)
The equation of normal is given by y – y1 = m(normal)(x – x1)
y - 1 = -1(x - (π/2) -1)
