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Question Number 49755 by ajfour last updated on 10/Dec/18

$$Solvesimultaneouslyfor\mathit{s}interms$$$$of\mathit{a}and\mathit{b}.$$$$h^2+{(b-k)}^2=s^2.....\left(i\right)$$$$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1.....\left(ii\right)$$$${(h-\frac{s}{2})}^2+(k+b\sqrt{1-\frac{s^2}{4a^2}})=s^2..\left(iii\right).$$

Answered by mr W last updated on 10/Dec/18

$$\left(ii\right)\times a^2:$$$$h^2+\frac{a^2}{b^2}{k}^2=a^2...\left(iv\right)$$$$\left(i\right)-\left(iv\right):$$$$b^2-2bk+k^2-\frac{a^2}{b^2}{k}^2=s^2-a^2$$$$(1-\frac{a^2}{b^2})k^2-2bk+(a^2+b^2-s^2)=0$$$$\Rightarrow k=\frac{b-b\sqrt{1-(1-\frac{a^2}{b^2})(1+\frac{a^2}{b^2}-\frac{s^2}{b^2})}}{(1-\frac{a^2}{b^2})}(only‘‘{-}^{″}?)$$$$let\lambda =\frac{a}{b},\delta =\frac{s}{b}$$$$\Rightarrow k=\frac{b-b\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)}$$$$\Rightarrow \frac{k}{b}=\frac{1-\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)}$$$$$$$$\left(i\right)-\left(iii\right):$$$$\frac{s}{2}(2h-\frac{s}{2})+(b-k+k+b\sqrt{1-\frac{s^2}{4a^2}})(b-k-k+b\sqrt{1-\frac{s^2}{4a^2}})=0$$$$\frac{s}{2}(2h-\frac{s}{2})+b(1+\sqrt{1-\frac{s^2}{4a^2}})(b+b\sqrt{1-\frac{s^2}{4a^2}}-2k)=0$$$$\frac{s}{2}(2h-\frac{s}{2})+b^2(1+\sqrt{1-\frac{s^2}{4a^2}})(1+\sqrt{1-\frac{s^2}{4a^2}}-\frac{2\pm 2\sqrt{1-(1-\frac{a^2}{b^2})(1+\frac{a^2}{b^2}-\frac{s^2}{b^2})}}{(1-\frac{a^2}{b^2})})=0$$$$s(h-\frac{s}{4})+b^2(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}})(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}}-\frac{2-2\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)})=0$$$$\frac{h}{a}=\frac{\delta }{4\lambda }-\frac{1}{\lambda \delta }(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}})(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}}-\frac{2-2\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)})=0$$$$\left(ii\right):$$$$\Rightarrow {\{\frac{\delta }{4\lambda }-\frac{1}{\lambda \delta }(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}})(1+\sqrt{1-\frac{{\delta }^2}{4{\lambda }^2}}-\frac{2-2\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)})\}}^2+{\left\{\frac{1-\sqrt{1-(1-{\lambda }^2)(1+{\lambda }^2-{\delta }^2)}}{(1-{\lambda }^2)}\right\}}^2=1$$$$solvefor\delta intermsof\lambda ....\left(onlynummericallypossible\right)$$

Commented by ajfour last updated on 10/Dec/18

$$ThankyouSir,Anybetteralternative$$$$forQ.49740?$$

Commented by mr W last updated on 10/Dec/18

$$nosir!Ihavenobetterway.$$

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