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Question-186833

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Question Number 186833 by mr W last updated on 11/Feb/23
Commented by mr W last updated on 11/Feb/23
as Q186762, but the well has the shape of an ellipse.
$${as}\:{Q}\mathrm{186762},\:{but}\:{the}\:{well}\:{has}\:{the}\:{shape} \\ $$$${of}\:{an}\:{ellipse}. \\ $$
Answered by mr W last updated on 12/Feb/23
Commented by mr W last updated on 11/Feb/23
AB=c=(√(a^2 +b^2 )) tan φ=(b/a) y_D =((ab)/c)=((ab)/( (√(a^2 +b^2 )))) x_D =(a^2 /c)−(c/2)=((a^2 −b^2 )/(2c))=((a^2 −b^2 )/(2(√(a^2 +b^2 )))) eqn. of big ellipse: (x^2 /u^2 )+(y^2 /v^2 )=1 (√(u^2 −v^2 ))=(c/2)=((√(a^2 +b^2 ))/2) ⇒u^2 −v^2 =((a^2 +b^2 )/4) ...(i) say C(u cos θ, v sin θ) on big ellipse and C(p cos ϕ, q sin ϕ) on small ellipse tan α=((v cos θ)/(u sin θ))=(v/(u tan θ)) tan β=((q cos ϕ)/(p sin ϕ))=(q/(p tan ϕ)) β−α=φ ((tan β−tan α)/(1+tan β tan α))=tan φ (((q/(p tan ϕ))−(v/(u tan θ)))/(1+(q/(p tan ϕ))×(v/(u tan θ))))=(b/a) ((qu tan θ−pv tan ϕ)/(pu tan θ tan ϕ+qv))=(b/a) tan ϕ=((q(au tan θ−bv))/(p(av+bu tan θ))) let m=tan θ ⇒tan ϕ=((q(aum−bv))/(p(av+bum))) u cos θ=x_D −p cos ϕ cos φ+q sin ϕ sin φ ⇒((u(√(a^2 +b^2 )))/( (√(m^2 +1))))=((a^2 −b^2 )/2)−pa cos ϕ+qb sin ϕ v sin θ=y_D −p cos ϕ sin φ−q sin ϕ cos φ ⇒((vm(√(a^2 +b^2 )))/( (√(m^2 +1))))=ab−pb cos ϕ−qa sin ϕ ⇒(b/2)+((bu−avm)/( (√((m^2 +1)(a^2 +b^2 )))))=q sin ϕ ⇒(a/2)−((au+bvm)/( (√((m^2 +1)(a^2 +b^2 )))))=p cos ϕ ⇒(b/2)+((bu−avm)/( (√((m^2 +1)(a^2 +b^2 )))))=(q^2 /( (√(p^2 (((av+bum)/(aum−bv)))^2 +q^2 )))) ...(ii) ⇒(a/2)−((au+bvm)/( (√((m^2 +1)(a^2 +b^2 )))))=(p^2 /( (√(p^2 +q^2 (((aum−bv)/(av+bum)))^2 )))) ...(iii) we can solve (i) to (iii) for u,v,m. example: a=10, b=6, p=3, q=2 ⇒u≈6.5261, v≈2.9309 shortest way length L_(min) =2u≈13.0522
$${AB}={c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\phi=\frac{{b}}{{a}} \\ $$$${y}_{{D}} =\frac{{ab}}{{c}}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${x}_{{D}} =\frac{{a}^{\mathrm{2}} }{{c}}−\frac{{c}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{c}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${eqn}.\:{of}\:{big}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{v}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\sqrt{{u}^{\mathrm{2}} −{v}^{\mathrm{2}} }=\frac{{c}}{\mathrm{2}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{v}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}}\:\:\:…\left({i}\right) \\ $$$${say}\:{C}\left({u}\:\mathrm{cos}\:\theta,\:{v}\:\mathrm{sin}\:\theta\right)\:{on}\:{big}\:{ellipse} \\ $$$${and}\:{C}\left({p}\:\mathrm{cos}\:\varphi,\:{q}\:\mathrm{sin}\:\varphi\right)\:{on}\:{small}\:{ellipse} \\ $$$$\mathrm{tan}\:\alpha=\frac{{v}\:\mathrm{cos}\:\theta}{{u}\:\mathrm{sin}\:\theta}=\frac{{v}}{{u}\:\mathrm{tan}\:\theta} \\ $$$$\mathrm{tan}\:\beta=\frac{{q}\:\mathrm{cos}\:\varphi}{{p}\:\mathrm{sin}\:\varphi}=\frac{{q}}{{p}\:\mathrm{tan}\:\varphi} \\ $$$$\beta−\alpha=\phi \\ $$$$\frac{\mathrm{tan}\:\beta−\mathrm{tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:\beta\:\mathrm{tan}\:\alpha}=\mathrm{tan}\:\phi \\ $$$$\frac{\frac{{q}}{{p}\:\mathrm{tan}\:\varphi}−\frac{{v}}{{u}\:\mathrm{tan}\:\theta}}{\mathrm{1}+\frac{{q}}{{p}\:\mathrm{tan}\:\varphi}×\frac{{v}}{{u}\:\mathrm{tan}\:\theta}}=\frac{{b}}{{a}} \\ $$$$\frac{{qu}\:\mathrm{tan}\:\theta−{pv}\:\mathrm{tan}\:\varphi}{{pu}\:\mathrm{tan}\:\theta\:\mathrm{tan}\:\varphi+{qv}}=\frac{{b}}{{a}} \\ $$$$\mathrm{tan}\:\varphi=\frac{{q}\left({au}\:\mathrm{tan}\:\theta−{bv}\right)}{{p}\left({av}+{bu}\:\mathrm{tan}\:\theta\right)} \\ $$$${let}\:{m}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=\frac{{q}\left({aum}−{bv}\right)}{{p}\left({av}+{bum}\right)} \\ $$$${u}\:\mathrm{cos}\:\theta={x}_{{D}} −{p}\:\mathrm{cos}\:\varphi\:\mathrm{cos}\:\phi+{q}\:\mathrm{sin}\:\varphi\:\mathrm{sin}\:\phi \\ $$$$\Rightarrow\frac{{u}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}−{pa}\:\mathrm{cos}\:\varphi+{qb}\:\mathrm{sin}\:\varphi \\ $$$${v}\:\mathrm{sin}\:\theta={y}_{{D}} −{p}\:\mathrm{cos}\:\varphi\:\mathrm{sin}\:\phi−{q}\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\phi \\ $$$$\Rightarrow\frac{{vm}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}={ab}−{pb}\:\mathrm{cos}\:\varphi−{qa}\:\mathrm{sin}\:\varphi \\ $$$$ \\ $$$$\Rightarrow\frac{{b}}{\mathrm{2}}+\frac{{bu}−{avm}}{\:\sqrt{\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}={q}\:\mathrm{sin}\:\varphi \\ $$$$\Rightarrow\frac{{a}}{\mathrm{2}}−\frac{{au}+{bvm}}{\:\sqrt{\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}={p}\:\mathrm{cos}\:\varphi \\ $$$$ \\ $$$$\Rightarrow\frac{{b}}{\mathrm{2}}+\frac{{bu}−{avm}}{\:\sqrt{\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}=\frac{{q}^{\mathrm{2}} }{\:\sqrt{{p}^{\mathrm{2}} \left(\frac{{av}+{bum}}{{aum}−{bv}}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} }}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\frac{{a}}{\mathrm{2}}−\frac{{au}+{bvm}}{\:\sqrt{\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}=\frac{{p}^{\mathrm{2}} }{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} \left(\frac{{aum}−{bv}}{{av}+{bum}}\right)^{\mathrm{2}} }}\:\:\:…\left({iii}\right) \\ $$$${we}\:{can}\:{solve}\:\left({i}\right)\:{to}\:\left({iii}\right)\:{for}\:{u},{v},{m}. \\ $$$$ \\ $$$$\boldsymbol{{example}}: \\ $$$${a}=\mathrm{10},\:{b}=\mathrm{6},\:{p}=\mathrm{3},\:{q}=\mathrm{2} \\ $$$$\Rightarrow{u}\approx\mathrm{6}.\mathrm{5261},\:{v}\approx\mathrm{2}.\mathrm{9309} \\ $$$${shortest}\:{way}\:{length}\:{L}_{{min}} =\mathrm{2}{u}\approx\mathrm{13}.\mathrm{0522} \\ $$
Commented by mr W last updated on 11/Feb/23

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