Question Number 186809 by ajfour last updated on 10/Feb/23
Answered by ajfour last updated on 10/Feb/23
Commented by ajfour last updated on 10/Feb/23
$${x}={p}=\mathrm{tan}\:\theta,\:{let}\:\:\mathrm{tan}\:\alpha={c} \\ $$$${y}={cx} \\ $$$${left}\:{corner}\:\:{A}\left(−{p}^{\mathrm{2}} ,{c}\right) \\ $$$$\bot\:{distance}\:{q} \\ $$$${q}\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right) \\ $$$${let}\:{sides}\:{ofcentral}\:{rectangle}\:{be} \\ $$$${s}\:{and}\:{t}. \\ $$$${s}={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)−{p}\mathrm{sin}\:\alpha \\ $$$$\:\:\:={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)−\frac{{cp}}{\:\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }} \\ $$$${t}=\left(\mathrm{1}+{p}^{\mathrm{2}} \right)\mathrm{cos}\:\alpha−\left({c}−\frac{\mathrm{1}}{{p}}\right)\mathrm{sin}\:\alpha \\ $$