Question Number 157169 by cortano last updated on 20/Oct/21
Answered by mr W last updated on 20/Oct/21
![x=r cos θ y=r sin θ r^2 cos^2 θ+2r^2 sin^2 θ+2r^2 cos θsin θ=2 r^2 (3+2sin 2θ−cos 2θ)=4 r^2 [3+(√5) sin (2θ−tan^(−1) (1/2))]=4 r=(2/( (√(3+(√5) sin (2θ−tan^(−1) (1/2)))))) r_(max) =(2/( (√(3−(√5)))))=maximum distance r_(min) =(2/( (√(3+(√5)))))=minimum distance](https://www.tinkutara.com/question/Q157194.png)
$${x}={r}\:\mathrm{cos}\:\theta \\ $$$${y}={r}\:\mathrm{sin}\:\theta \\ $$$${r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{r}^{\mathrm{2}} \mathrm{cos}\:\theta\mathrm{sin}\:\theta=\mathrm{2} \\ $$$${r}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2sin}\:\mathrm{2}\theta−\mathrm{cos}\:\mathrm{2}\theta\right)=\mathrm{4} \\ $$$${r}^{\mathrm{2}} \left[\mathrm{3}+\sqrt{\mathrm{5}}\:\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)\right]=\mathrm{4} \\ $$$${r}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}\:\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)}} \\ $$$${r}_{{max}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}}={maximum}\:{distance} \\ $$$${r}_{{min}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}}={minimum}\:{distance} \\ $$
Commented by cortano last updated on 21/Oct/21
$${yes}\:{bravo} \\ $$