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प्रश्न
x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ
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उत्तर
Given that: x = 3 cosθ – 2 cos3θ and y = 3sinθ – 2 sin3θ.
Differentiating both the parametric functions w.r.t. θ
`"dx"/("d"theta) = -3 sin theta - 6cos^2theta * "d"/("d"theta) (cos theta)`
= – 3 sin θ – 6 cos2θ . (– sin θ)
= – 3 sin θ + 6 cos2θ . sin θ
`"dy"/("d"theta) = 3 os theta - 6 sin^2theta * "d"/("d"theta) (sin theta)`
= = 3 cos θ – 6 sin2θ . cos θ
∴ `"dy"/"dx" = ("dy"/("d"theta))/("dx"/("d"theta))`
= `(3 cos theta - 6 sin^2theta cos theta)/(-3sin theta + 6cos^2 theta * sin theta)`
⇒ `"dy"/"dx" = (cos theta (3 - 6sin^2theta))/(sintheta(-3 + 6 cos^2 theta))`
= `(costheta[3 - 6(1 - cos^2theta)])/(sintheta[-3 + 6cos^2theta])`
= `cot theta ((3 - 6 + 6 cos^2 theta)/(-3 + 6 cos^2theta))`
= `cot theta ((-3 + 6 cos^2theta)/(-3 + 6 cos^2 theta))`
= cot θ
∴ `"dy"/"dx"` = cot θ.
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