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Question-199413

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Question Number 199413 by universe last updated on 03/Nov/23
Answered by ajfour last updated on 03/Nov/23
bcos α=R {Rtan α+((a/b))R}^2 +{((a/b))Rtan α}^2 =R^2 say tan α=t ⇒ t^2 +(a^2 /b^2 )+((2at)/b)+(a^2 /b^2 )t^2 =1 (a^2 +b^2 )t^2 +2abt+a^2 −b^2 =0 t=tan α=(((√(a^2 b^2 +(b^4 −a^4 )))−ab)/(a^2 +b^2 ))
$${b}\mathrm{cos}\:\alpha={R} \\ $$$$\left\{{R}\mathrm{tan}\:\alpha+\left(\frac{{a}}{{b}}\right){R}\right\}^{\mathrm{2}} +\left\{\left(\frac{{a}}{{b}}\right){R}\mathrm{tan}\:\alpha\right\}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${say}\:\:\:\mathrm{tan}\:\alpha={t} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{\mathrm{2}{at}}{{b}}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{t}^{\mathrm{2}} =\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\mathrm{2}{abt}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\alpha=\frac{\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({b}^{\mathrm{4}} −{a}^{\mathrm{4}} \right)}−{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$
Answered by mr W last updated on 03/Nov/23
R=b cos α R^2 =a^2 +(b sin α)^2 −2ab sin α cos ((π/2)+(π/2)−α) b^2 cos^2 α=a^2 +b^2 sin^2 α+ab sin 2α b^2 cos 2α−ab sin 2α=a^2 let t=tan α b^2 ×((1−t^2 )/(1+t^2 ))−ab×((2t)/(1+t^2 ))=a^2 (a^2 +b^2 )t^2 +2abt+a^2 −b^2 =0 ⇒t=(((√(a^2 b^2 +b^4 −a^4 ))−ab)/(a^2 +b^2 ))=tan α
$${R}={b}\:\mathrm{cos}\:\alpha \\ $$$${R}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({b}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$${b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha+{ab}\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$${b}^{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\alpha−{ab}\:\mathrm{sin}\:\mathrm{2}\alpha={a}^{\mathrm{2}} \\ $$$${let}\:{t}=\mathrm{tan}\:\alpha \\ $$$${b}^{\mathrm{2}} ×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }−{ab}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }={a}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\mathrm{2}{abt}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} }−{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{tan}\:\alpha \\ $$

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