Question

Show that the equation of normal at any point on the curve x = 3cos θ – cos³θ, y = 3sinθ – sin³θ is 4 (y cos³θ – x sin³θ) = 3 sin 4θ

Answer 1

We have x = 3cos θ – cos³θ

Therefore, `"dx"/("d"theta)` = –3sin θ + 3cos²θ sinθ

= – 3sinθ (1 – cos²θ)

= –3sin³θ .

`"dy"/("d"theta) = - (cos^3theta)/(sin^3theta)`.

Therefore, slope of normal = `+ (sin^3theta)/(cos^2theta)`

Hence the equation of normal is

y – (3sinθ – sin³θ) = `(sin^3theta)/(cos^2theta)` [x – (3cosθ – cos³θ)]

⇒ y cos³θ – 3sinθ cos³θ + sin³θ cos³θ = xsin³θ – 3sin³θ cosθ + sin³θ cos³θ

⇒ y cos³θ – xsin³θ = 3sinθ cosθ (cos²θ – sin²θ)

= `3/2 sin2theta * cos2theta`

= `3/4 sin4theta`

or 4 (y cos³θ – x sin³θ) = 3 sin4θ.