Question Number 31820 by mondodotto@gmail.com last updated on 15/Mar/18
Answered by mrW2 last updated on 15/Mar/18
$$\left({i}\right) \\ $$$$\frac{{dx}}{{d}\theta}=−\mathrm{4}\:\mathrm{sin}\:\theta \\ $$$$\frac{{dy}}{{d}\theta}=\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\theta}×\frac{\mathrm{1}}{\frac{{dx}}{{d}\theta}}=−\frac{\mathrm{3}\:\mathrm{cos}\:\theta}{\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\left({ii}\right) \\ $$$$\left({y}−\mathrm{3sin}\:\theta\right)=\frac{\mathrm{4}\:\mathrm{sin}\:\theta}{\mathrm{3}\:\mathrm{cos}\:\theta}\left({x}−\mathrm{4cos}\:\theta\right) \\ $$$$\left({iii}\right) \\ $$$$\left({y}−\mathrm{3sin}\:\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{4}}{\mathrm{3}}\left({x}−\mathrm{4cos}\:\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left({y}−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\mathrm{4}}{\mathrm{3}}\left({x}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}{x}}{\mathrm{3}}−\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$$\left({iv}\right) \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{2}},\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$