मराठी

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 - Mathematics

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प्रश्न

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

बेरीज
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उत्तर

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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पाठ 6: Application Of Derivatives - Solved Examples [पृष्ठ १२७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 6 Application Of Derivatives
Solved Examples | Q 14 | पृष्ठ १२७

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