Question Number 80731 by ajfour last updated on 05/Feb/20
Commented by ajfour last updated on 05/Feb/20
$${The}\:{pentagon}\:{has}\:{each}\:{side}\:{s}. \\ $$$${Find}\:{s},\:{in}\:{terms}\:{of}\:{ellipse} \\ $$$${parameters}\:{a}\:{and}\:{b}. \\ $$
Answered by mr W last updated on 06/Feb/20
![A(a cos α, b sin α) B(a cos β, −b sin β) (a cos α)^2 +(b−b sin α)^2 =s^2 (a cos α−a cos β)^2 +(b sin α+b sin β)^2 =s^2 a cos β=(s/2) let (b/a)=μ, (s/a)=λ cos^2 α+μ^2 (1−sin α)^2 =λ^2 ...(i) (cos α−cos β)^2 +μ^2 (sin α+sin β)^2 =λ^2 ...(ii) cos β=(λ/2) ...(iii) (1−μ^2 )sin^2 α+2μ^2 sin α−(1+μ^2 −λ^2 )=0 if μ=1: sin α=1−(λ^2 /2) if μ<1: ⇒sin α=((μ^2 −(√(1−(1−μ^2 )λ^2 )))/(1−μ^2 )) ⇒α=sin^(−1) [((μ^2 −(√(1−(1−μ^2 )λ^2 )))/(1−μ^2 ))] ⇒β=cos^(−1) (λ/2) put them into (i) to get λ. examples: μ=(b/a)=1: λ=(s/a)=1.1756 μ=(b/a)=(2/3): λ=(s/a)=1.3312](https://www.tinkutara.com/question/Q80740.png)
$${A}\left({a}\:\mathrm{cos}\:\alpha,\:{b}\:\mathrm{sin}\:\alpha\right) \\ $$$${B}\left({a}\:\mathrm{cos}\:\beta,\:−{b}\:\mathrm{sin}\:\beta\right) \\ $$$$\left({a}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left({b}−{b}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$$\left({a}\:\mathrm{cos}\:\alpha−{a}\:\mathrm{cos}\:\beta\right)^{\mathrm{2}} +\left({b}\:\mathrm{sin}\:\alpha+{b}\:\mathrm{sin}\:\beta\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${a}\:\mathrm{cos}\:\beta=\frac{{s}}{\mathrm{2}} \\ $$$${let}\:\frac{{b}}{{a}}=\mu,\:\frac{{s}}{{a}}=\lambda \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha+\mu^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} =\lambda^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta\right)^{\mathrm{2}} +\mu^{\mathrm{2}} \left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)^{\mathrm{2}} =\lambda^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\mathrm{cos}\:\beta=\frac{\lambda}{\mathrm{2}}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{sin}^{\mathrm{2}} \:\alpha+\mathrm{2}\mu^{\mathrm{2}} \:\mathrm{sin}\:\alpha−\left(\mathrm{1}+\mu^{\mathrm{2}} −\lambda^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${if}\:\mu=\mathrm{1}: \\ $$$$\mathrm{sin}\:\alpha=\mathrm{1}−\frac{\lambda^{\mathrm{2}} }{\mathrm{2}} \\ $$$${if}\:\mu<\mathrm{1}: \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mu^{\mathrm{2}} −\sqrt{\mathrm{1}−\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }}{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$$\Rightarrow\alpha=\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mu^{\mathrm{2}} −\sqrt{\mathrm{1}−\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\lambda^{\mathrm{2}} }}{\mathrm{1}−\mu^{\mathrm{2}} }\right] \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\lambda}{\mathrm{2}} \\ $$$${put}\:{them}\:{into}\:\left({i}\right)\:{to}\:{get}\:\lambda. \\ $$$$ \\ $$$${examples}: \\ $$$$\mu=\frac{{b}}{{a}}=\mathrm{1}:\:\lambda=\frac{{s}}{{a}}=\mathrm{1}.\mathrm{1756} \\ $$$$\mu=\frac{{b}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}}:\:\lambda=\frac{{s}}{{a}}=\mathrm{1}.\mathrm{3312} \\ $$
Commented by ajfour last updated on 06/Feb/20
$${let}\:{me}\:{some}\:{time}\:{to}\:{follow},\:{Sir}. \\ $$