If x=acos3θ,y=bsin3θ, then d2ydx2=

x=acos3θdxdθ=−3acos2θ.sinθy=bsin3θdydθ=3bsin2θ.cosθdydθ=−batanθ [∴tanθ=(aybx)13]dydx=−ba(aybx)13=(−ba)23.(yx)13d2ydx2=(−ba)23.13.(yx)−23.(xdydx−y)x2=(−ba)23.13.(bsin3θacos3θ)−23(acos3θ(−ba)23.(bsin3θacos3θ)13−bsin3θ](acos3θ)2=(−ba)23.13(ba)−23.cos2θsin2θ[acos3θ.(−ba)23.(ba)13sinθ−bsin3θcosθa2cos6θ]=−13a2×cos2θsin2θ×1cos6θ[−acos3θ(ba)sinθcosθ−bsin3θ]=−13a2.cos2θsin2θ×1cos6θ[−bcos2θ.sinθ−bsin3θ]=−13a2[1sin2θ.1cos4θ[−b(1−sin2θ)sinθ−bsin3θ])=13a2cosec2θ.sec4θ1(bsinθ−bsin3θ+bsin3θ]=b3a2cosecθ.sec4θShort method:x=acos3θ⇒dxdθ=−3acos2θ.sinθy=bsin3θ⇒dydθ=3bsin2θ.cosθdydx=−batanθd2ydx2=−basec2θ.dθdx=−basec21−3acos2θ.sinθ=b3a2cosecθ.sec4θ

If x=acos3θ,y=bsin3θ, then d2ydx2=

x=acos3θdxdθ=−3acos2θ.sinθy=bsin3θdydθ=3bsin2θ.cosθdydθ=−batanθ [∴tanθ=(aybx)13]dydx=−ba(aybx)13=(−ba)23.(yx)13d2ydx2=(−ba)23.13.(yx)−23.(xdydx−y)x2=(−ba)23.13.(bsin3θacos3θ)−23(acos3θ(−ba)23.(bsin3θacos3θ)13−bsin3θ](acos3θ)2=(−ba)23.13(ba)−23.cos2θsin2θ[acos3θ.(−ba)23.(ba)13sinθ−bsin3θcosθa2cos6θ]=−13a2×cos2θsin2θ×1cos6θ[−acos3θ(ba)sinθcosθ−bsin3θ]=−13a2.cos2θsin2θ×1cos6θ[−bcos2θ.sinθ−bsin3θ]=−13a2[1sin2θ.1cos4θ[−b(1−sin2θ)sinθ−bsin3θ])=13a2cosec2θ.sec4θ1(bsinθ−bsin3θ+bsin3θ]=b3a2cosecθ.sec4θShort method:x=acos3θ⇒dxdθ=−3acos2θ.sinθy=bsin3θ⇒dydθ=3bsin2θ.cosθdydx=−batanθd2ydx2=−basec2θ.dθdx=−basec21−3acos2θ.sinθ=b3a2cosecθ.sec4θ

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If $x = a \cos^{3} θ, y = b \sin^{3} θ$ , then $\frac{d^{2} y}{d x^{2}} =$

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$\frac{b}{3 a^{2}} \sec^{2} θ \cos e c^{2} θ$

$\frac{b}{3 a^{2}} \cos e c^{4} θ \sec θ$

$\frac{b}{3 a^{2}} \sec θ \cos e c^{4} θ$

$\frac{b}{3 a^{2}} \sec^{4} θ \cos e c θ$

answer is A.

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Detailed Solution

$x = a \cos^{3} θ$

$\frac{d x}{d θ} = - 3 a \cos^{2} θ . \sin θ$

$y = b \sin^{3} θ$

$\frac{d y}{d θ} = 3 b \sin^{2} θ . \cos θ$

$\frac{d y}{d θ} = \frac{- b}{a} \tan θ [∴ \tan θ = {(\frac{a y}{b x})}^{\frac{1}{3}}]$

$\frac{d y}{d x} = \frac{- b}{a} {(\frac{a y}{b x})}^{\frac{1}{3}} = {(\frac{- b}{a})}^{\frac{2}{3}} . {(\frac{y}{x})}^{\frac{1}{3}}$

$\frac{d^{2} y}{d x^{2}} = {(- \frac{b}{a})}^{\frac{2}{3}} . \frac{1}{3} . {(\frac{y}{x})}^{\frac{- 2}{3}} . \frac{(\frac{x d y}{d x} - y)}{x^{2}}$

$= {(\frac{- b}{a})}^{\frac{2}{3}} . \frac{1}{3} . {(\frac{b \sin^{3} θ}{a \cos^{3} θ})}^{\frac{- 2}{3}} \frac{(a \cos^{3} θ {(\frac{- b}{a})}^{\frac{2}{3}} . {(\frac{b \sin^{3} θ}{a \cos^{3} θ})}^{\frac{1}{3}} - b \sin^{3} θ]}{{(a \cos^{3} θ)}^{2}}$

$= {(\frac{- b}{a})}^{\frac{2}{3}} . \frac{1}{3} {(\frac{b}{a})}^{\frac{- 2}{3}} . \frac{\cos^{2} θ}{\sin^{2} θ} [\frac{a \cos^{3} θ . {(\frac{- b}{a})}^{\frac{2}{3}} . {(\frac{b}{a})}^{\frac{1}{3}} \sin θ - b \sin^{3} θ}{\cos θ a^{2} \cos^{6} θ}]$

$= \frac{- 1}{3 a^{2}} \times \frac{\cos^{2} θ}{\sin^{2} θ} \times \frac{1}{\cos^{6} θ} [- a \cos^{3} θ (\frac{b}{a}) \frac{\sin θ}{\cos θ} - b \sin^{3} θ]$