Question Number 130795 by mr W last updated on 29/Jan/21 | ||
Commented by mr W last updated on 29/Jan/21 | ||
$${find}\:{the}\:{locus}\:{of}\:{point}\:{P}. \\ $$ | ||
Answered by mr W last updated on 29/Jan/21 | ||
Commented by mr W last updated on 29/Jan/21 | ||
$$\mu=\frac{{b}}{{a}} \\ $$$${say}\:{A}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=\frac{{b}\:\mathrm{cos}\:\theta}{{a}\:\mathrm{sin}\:\theta}=\frac{\mu}{\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mu}{\mathrm{tan}\:\theta}\right) \\ $$$${x}_{{P}} ={a}\:\mathrm{cos}\:\theta−{a}\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\mathrm{2}\varphi+{b}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\mathrm{2}\varphi \\ $$$$\Rightarrow{x}_{{P}} ={a}\left[\mathrm{cos}\:\theta\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\varphi\right)+\mu\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\mathrm{2}\varphi\right] \\ $$$${y}_{{P}} ={b}\:\mathrm{sin}\:\theta+{a}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}\varphi+{b}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\:\mathrm{cos}\:\mathrm{2}\varphi \\ $$$$\Rightarrow{y}_{{P}} ={a}\left[\mu\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}\varphi+\mu\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\:\mathrm{cos}\:\mathrm{2}\varphi\right] \\ $$ | ||
Commented by mr W last updated on 29/Jan/21 | ||
Commented by mr W last updated on 29/Jan/21 | ||
Commented by mr W last updated on 29/Jan/21 | ||