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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

$$asQ186762,butthewellhastheshape$$$$ofanellipse.$$

Answered by mr W last updated on 12/Feb/23

Commented by mr W last updated on 11/Feb/23

$$AB=c=\sqrt{a^2+b^2}$$$$\tan \varphi =\frac{b}{a}$$$$y_D=\frac{ab}{c}=\frac{ab}{\sqrt{a^2+b^2}}$$$$x_D=\frac{a^2}{c}-\frac{c}{2}=\frac{a^2-b^2}{2c}=\frac{a^2-b^2}{2\sqrt{a^2+b^2}}$$$$eqn.ofbigellipse:$$$$\frac{x^2}{u^2}+\frac{y^2}{v^2}=1$$$$\sqrt{u^2-v^2}=\frac{c}{2}=\frac{\sqrt{a^2+b^2}}{2}$$$$\Rightarrow u^2-v^2=\frac{a^2+b^2}{4}...\left(i\right)$$$$sayC(u\cos \theta ,v\sin \theta )onbigellipse$$$$andC(p\cos \phi ,q\sin \phi )onsmallellipse$$$$\tan \alpha =\frac{v\cos \theta }{u\sin \theta }=\frac{v}{u\tan \theta }$$$$\tan \beta =\frac{q\cos \phi }{p\sin \phi }=\frac{q}{p\tan \phi }$$$$\beta -\alpha =\varphi$$$$\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\tan \varphi$$$$\frac{\frac{q}{p\tan \phi }-\frac{v}{u\tan \theta }}{1+\frac{q}{p\tan \phi }\times \frac{v}{u\tan \theta }}=\frac{b}{a}$$$$\frac{qu\tan \theta -pv\tan \phi }{pu\tan \theta \tan \phi +qv}=\frac{b}{a}$$$$\tan \phi =\frac{q(au\tan \theta -bv)}{p(av+bu\tan \theta )}$$$$letm=\tan \theta$$$$\Rightarrow \tan \phi =\frac{q(aum-bv)}{p(av+bum)}$$$$u\cos \theta =x_D-p\cos \phi \cos \varphi +q\sin \phi \sin \varphi$$$$\Rightarrow \frac{u\sqrt{a^2+b^2}}{\sqrt{m^2+1}}=\frac{a^2-b^2}{2}-pa\cos \phi +qb\sin \phi$$$$v\sin \theta =y_D-p\cos \phi \sin \varphi -q\sin \phi \cos \varphi$$$$\Rightarrow \frac{vm\sqrt{a^2+b^2}}{\sqrt{m^2+1}}=ab-pb\cos \phi -qa\sin \phi$$$$$$$$\Rightarrow \frac{b}{2}+\frac{bu-avm}{\sqrt{(m^2+1)(a^2+b^2)}}=q\sin \phi$$$$\Rightarrow \frac{a}{2}-\frac{au+bvm}{\sqrt{(m^2+1)(a^2+b^2)}}=p\cos \phi$$$$$$$$\Rightarrow \frac{b}{2}+\frac{bu-avm}{\sqrt{(m^2+1)(a^2+b^2)}}=\frac{q^2}{\sqrt{p^2{\left(\frac{av+bum}{aum-bv}\right)}^2+q^2}}...\left(ii\right)$$$$\Rightarrow \frac{a}{2}-\frac{au+bvm}{\sqrt{(m^2+1)(a^2+b^2)}}=\frac{p^2}{\sqrt{p^2+q^2{\left(\frac{aum-bv}{av+bum}\right)}^2}}...\left(iii\right)$$$$wecansolve\left(i\right)to\left(iii\right)foru,v,m.$$$$$$$$\mathit{e}\mathit{x}\mathit{a}\mathit{m}\mathit{p}\mathit{l}\mathit{e}:$$$$a=10,b=6,p=3,q=2$$$$\Rightarrow u\approx 6.5261,v\approx 2.9309$$$$shortestwaylength{L}_{min}=2u\approx 13.0522$$

Commented by mr W last updated on 11/Feb/23

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