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Question Number 98988 by Rio Michael last updated on 17/Jun/20

Find the tangent at the poles for the polar equation r = a sin 2θ.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{for}\:\mathrm{the}\:\mathrm{polar} \\ $$$$\mathrm{equation}\:{r}\:=\:{a}\:\mathrm{sin}\:\mathrm{2}\theta. \\ $$

Answered by mr W last updated on 17/Jun/20

x=r cos θ=a sin 2θ cos θ y=r sin θ=a sin 2θ sin θ (dy/dx)=(dy/dθ)×(1/(dx/dθ))=((2 cos 2θ sin θ+sin 2θ cos θ)/(2 cos 2θ cos θ−sin 2θ sin θ)) =((2 tan θ+tan 2θ)/(2−tan θ tan 2θ))

$${x}={r}\:\mathrm{cos}\:\theta={a}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta \\ $$$${y}={r}\:\mathrm{sin}\:\theta={a}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\theta}×\frac{\mathrm{1}}{\frac{{dx}}{{d}\theta}}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta+\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta}{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta−\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\theta} \\ $$$$=\frac{\mathrm{2}\:\mathrm{tan}\:\theta+\mathrm{tan}\:\mathrm{2}\theta}{\mathrm{2}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\mathrm{2}\theta} \\ $$

Commented by Rio Michael last updated on 17/Jun/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 17/Jun/20

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