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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
Solution
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.
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