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

Solve simultaneously for s in terms of a and b. h^2 +(b−k)^2 = s^2 .....(i) (h^2 /a^2 )+(k^2 /b^2 ) = 1 .....(ii) (h−(s/2))^2 +(k+b(√(1−(s^2 /(4a^2 )))) )= s^2 ..(iii).

$${Solve}\:{simultaneously}\:{for}\:\boldsymbol{{s}}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$$${h}^{\mathrm{2}} +\left({b}−{k}\right)^{\mathrm{2}} =\:{s}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({ii}\right) \\ $$$$\left({h}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({k}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\:\right)=\:{s}^{\mathrm{2}} \:\:\:..\left({iii}\right). \\ $$

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

(ii)×a^2 : h^2 +(a^2 /b^2 )k^2 =a^2 ...(iv) (i)−(iv): b^2 −2bk+k^2 −(a^2 /b^2 )k^2 =s^2 −a^2 (1−(a^2 /b^2 ))k^2 −2bk+(a^2 +b^2 −s^2 )=0 ⇒k=((b−b(√(1−(1−(a^2 /b^2 ))(1+(a^2 /b^2 )−(s^2 /b^2 )))))/((1−(a^2 /b^2 )))) (only “−”?) let λ=(a/b),δ=(s/b) ⇒k=((b−b(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 ))) ⇒(k/b)=((1−(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 ))) (i)−(iii): (s/2)(2h−(s/2))+(b−k+k+b(√(1−(s^2 /(4a^2 )))))(b−k−k+b(√(1−(s^2 /(4a^2 )))))=0 (s/2)(2h−(s/2))+b(1+(√(1−(s^2 /(4a^2 )))))(b+b(√(1−(s^2 /(4a^2 ))))−2k)=0 (s/2)(2h−(s/2))+b^2 (1+(√(1−(s^2 /(4a^2 )))))(1+(√(1−(s^2 /(4a^2 ))))−((2±2(√(1−(1−(a^2 /b^2 ))(1+(a^2 /b^2 )−(s^2 /b^2 )))))/((1−(a^2 /b^2 )))))=0 s(h−(s/4))+b^2 (1+(√(1−(δ^2 /(4λ^2 )))))(1+(√(1−(δ^2 /(4λ^2 ))))−((2−2(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 ))))=0 (h/a)=(δ/(4λ))−(1/(λδ))(1+(√(1−(δ^2 /(4λ^2 )))))(1+(√(1−(δ^2 /(4λ^2 ))))−((2−2(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 ))))=0 (ii): ⇒{(δ/(4λ))−(1/(λδ))(1+(√(1−(δ^2 /(4λ^2 )))))(1+(√(1−(δ^2 /(4λ^2 ))))−((2−2(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 ))))}^2 +{((1−(√(1−(1−λ^2 )(1+λ^2 −δ^2 ))))/((1−λ^2 )))}^2 =1 solve for δ in terms of λ.... (only nummerically possible)

$$\left({ii}\right)×{a}^{\mathrm{2}} : \\ $$$${h}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{k}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:\:...\left({iv}\right) \\ $$$$\left({i}\right)−\left({iv}\right): \\ $$$${b}^{\mathrm{2}} −\mathrm{2}{bk}+{k}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{k}^{\mathrm{2}} ={s}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){k}^{\mathrm{2}} −\mathrm{2}{bk}+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{{b}−{b}\sqrt{\mathrm{1}−\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{s}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}}{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}\:\:\:\:\:\:\:\:\left({only}\:``−''?\right) \\ $$$${let}\:\lambda=\frac{{a}}{{b}},\delta=\frac{{s}}{{b}} \\ $$$$\Rightarrow{k}=\frac{{b}−{b}\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\frac{{k}}{{b}}=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\left({i}\right)−\left({iii}\right): \\ $$$$\frac{{s}}{\mathrm{2}}\left(\mathrm{2}{h}−\frac{{s}}{\mathrm{2}}\right)+\left({b}−{k}+{k}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\right)\left({b}−{k}−{k}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\right)=\mathrm{0} \\ $$$$\frac{{s}}{\mathrm{2}}\left(\mathrm{2}{h}−\frac{{s}}{\mathrm{2}}\right)+{b}\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\right)\left({b}+{b}\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}−\mathrm{2}{k}\right)=\mathrm{0} \\ $$$$\frac{{s}}{\mathrm{2}}\left(\mathrm{2}{h}−\frac{{s}}{\mathrm{2}}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}−\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{1}−\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{s}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}}{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}\right)=\mathrm{0} \\ $$$${s}\left({h}−\frac{{s}}{\mathrm{4}}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}−\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)}\right)=\mathrm{0} \\ $$$$\frac{{h}}{{a}}=\frac{\delta}{\mathrm{4}\lambda}−\frac{\mathrm{1}}{\lambda\delta}\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}−\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)}\right)=\mathrm{0} \\ $$$$\left({ii}\right): \\ $$$$\Rightarrow\left\{\frac{\delta}{\mathrm{4}\lambda}−\frac{\mathrm{1}}{\lambda\delta}\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\delta^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} }}−\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)}\right)\right\}^{\mathrm{2}} +\left\{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\lambda^{\mathrm{2}} −\delta^{\mathrm{2}} \right)}}{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)}\right\}^{\mathrm{2}} =\mathrm{1} \\ $$$${solve}\:{for}\:\delta\:{in}\:{terms}\:{of}\:\lambda....\:\left({only}\:{nummerically}\:{possible}\right) \\ $$

Commented by ajfour last updated on 10/Dec/18

Thank you Sir, Any better alternative for Q.49740 ?

$${Thank}\:{you}\:{Sir},\:{Any}\:{better}\:{alternative} \\ $$$${for}\:{Q}.\mathrm{49740}\:? \\ $$

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

no sir! I have no better way.

$${no}\:{sir}!\:{I}\:{have}\:{no}\:{better}\:{way}. \\ $$

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