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Question-186809




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=tan θ, let  tan α=c  y=cx  left corner  A(−p^2 ,c)  ⊥ distance q  q(√(1+c^2 ))=c(1+p^2 )  let sides ofcentral rectangle be  s and t.  s=c(1+p^2 )−psin α     =c(1+p^2 )−((cp)/( (√(1+c^2 ))))  t=(1+p^2 )cos α−(c−(1/p))sin α
$${x}={p}=\mathrm{tan}\:\theta,\:{let}\:\:\mathrm{tan}\:\alpha={c} \\ $$$${y}={cx} \\ $$$${left}\:{corner}\:\:{A}\left(−{p}^{\mathrm{2}} ,{c}\right) \\ $$$$\bot\:{distance}\:{q} \\ $$$${q}\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right) \\ $$$${let}\:{sides}\:{ofcentral}\:{rectangle}\:{be} \\ $$$${s}\:{and}\:{t}. \\ $$$${s}={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)−{p}\mathrm{sin}\:\alpha \\ $$$$\:\:\:={c}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)−\frac{{cp}}{\:\sqrt{\mathrm{1}+{c}^{\mathrm{2}} }} \\ $$$${t}=\left(\mathrm{1}+{p}^{\mathrm{2}} \right)\mathrm{cos}\:\alpha−\left({c}−\frac{\mathrm{1}}{{p}}\right)\mathrm{sin}\:\alpha \\ $$

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