Show that the equation of normal at any point t on the curve and is

Given: x = 3 cos t – cos³ t and y = 3 sin t – sin³ t

To prove: the Equation of normal at any point t on the curve is

4(y cos³ t – x sin³ t) = 3 sin 4t

Formula used:

The equation of a line with slope m and a point (a, b) is given by,

(y – b) = m(x – a)

If the slope of tangent and normal is m₁ and m₂ respectively,

m₁ × m₂ = -1

The slope of the tangent of a line is given by

Here x = 3 cos t – cos³ t and y = 3 sin t – sin³ t

x = 3 cos t – cos³ t

Differentiating both sides with respect to t:

y = 3 sin t – sin³ t

Differentiating both sides with respect to t:

The slope of tangent × Slope of normal = -1

The equation of normal,

⇒ y cos³ t – 3 sin t cos³ t + cos³ t sin³ t = x sin³ t – 3 cos t sin³ t + cos³ t sin³ t

⇒ y cos³ t – x sin³ t = 3 sin t cos³ t – 3 cos t sin³ t

⇒ y cos³ t – x sin³ t = 3 sin t cos t(cos² t – sin² t)

{∵ cos² t – sin² t = cos 2t}

{∵ sin 2t = 2 sin t cos t}

⇒ 4(y cos³ t – x sin³ t) = 3 sin 4t

Hence proved