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NCERT Exemplar · Q50

Q.Equation of the line passing through the point (acos⁡3θ, asin⁡3θ)(a\cos^3\theta,\ a\sin^3\theta) and perpendicular to the line xsec⁡θ+y cosec θ=ax\sec\theta + y\,\mathrm{cosec}\,\theta = a is xcos⁡θ−ysin⁡θ=asin⁡2θx\cos\theta - y\sin\theta = a\sin 2\theta.

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The perpendicular line through (acos⁡3θ, asin⁡3θ)(a\cos^3\theta,\,a\sin^3\theta) simplifies to xcos⁡θ−ysin⁡θ=acos⁡2θx\cos\theta-y\sin\theta=a\cos2\theta — so the given statement, which claims the right-hand side is asin⁡2θa\sin2\theta, is not correct in general.

Step 1: Slope of the given line

xsec⁡θ+ycsc⁡θ=a  ⟹  xcos⁡θ+ysin⁡θ=ax\sec\theta+y\csc\theta=a \implies \frac{x}{\cos\theta}+\frac{y}{\sin\theta}=a

Multiply through by sin⁡θcos⁡θ\sin\theta\cos\theta:

xsin⁡θ+ycos⁡θ=asin⁡θcos⁡θx\sin\theta+y\cos\theta=a\sin\theta\cos\theta

Solving for yy: y=asin⁡θcos⁡θ−xsin⁡θcos⁡θ=asin⁡θ−xtan⁡θy=\dfrac{a\sin\theta\cos\theta-x\sin\theta}{\cos\theta}=a\sin\theta-x\tan\theta, so its slope is m1=−tan⁡θm_1=-\tan\theta.

Step 2: Slope of the perpendicular line

m2=−1m1=1tan⁡θ=cot⁡θm_2=-\frac{1}{m_1}=\frac{1}{\tan\theta}=\cot\theta

Step 3: Equation through (acos⁡3θ, asin⁡3θ)(a\cos^3\theta,\ a\sin^3\theta)

y−asin⁡3θ=cot⁡θ (x−acos⁡3θ)y-a\sin^3\theta=\cot\theta\,(x-a\cos^3\theta)

Multiply both sides by sin⁡θ\sin\theta:

ysin⁡θ−asin⁡4θ=xcos⁡θ−acos⁡4θy\sin\theta-a\sin^4\theta=x\cos\theta-a\cos^4\theta

Rearrange:

xcos⁡θ−ysin⁡θ=a(cos⁡4θ−sin⁡4θ)=a(cos⁡2θ−sin⁡2θ)(cos⁡2θ+sin⁡2θ)=acos⁡2θx\cos\theta-y\sin\theta=a\left(\cos^4\theta-\sin^4\theta\right)=a(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)=a\cos2\theta …

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