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Question Number 53492 by ajfour last updated on 22/Jan/19

Commented by ajfour last updated on 22/Jan/19

$$Findparameters\mathit{a}and\mathit{b}ofan$$$$ellipsethatcircumscribesan$$$$equilateraltriangleofside\mathit{t}and$$$$evenasquareofside\mathit{s}.$$

Answered by mr W last updated on 22/Jan/19

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{s^2}{4a^2}+\frac{s^2}{4b^2}=1$$$$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{4}{s^2}$$$$\frac{t^2}{4a^2}+\frac{{(\frac{\sqrt{3}t}{2}-b)}^2}{b^2}=1$$$$\Rightarrow \frac{t^2}{a^2}+\frac{{(\sqrt{3}t-2b)}^2}{b^2}=4$$$$\Rightarrow \frac{4t^2}{s^2}-\frac{t^2}{b^2}+{(\sqrt{3}\frac{t}{b}-2)}^2=4$$$$with\mu =\frac{t}{s},\lambda =\frac{t}{b}$$$$\Rightarrow 4{\mu }^2-{\lambda }^2+{(\sqrt{3}\lambda -2)}^2=4$$$$\Rightarrow {\lambda }^2-2\sqrt{3}\lambda +2{\mu }^2=0$$$$\Rightarrow \lambda =\frac{t}{b}=\sqrt{3}+\sqrt{3-2{\mu }^2}$$$$\Rightarrow \frac{s}{b}\times \mu =\sqrt{3}+\sqrt{3-2{\mu }^2}$$$$\Rightarrow b=\frac{t}{\sqrt{3}+\sqrt{3-2{\mu }^2}}$$$$\Rightarrow a=\frac{s}{\sqrt{4-{\left(\frac{s}{b}\right)}^2}}=\frac{s}{\sqrt{4-{\left(\frac{\sqrt{3}+\sqrt{3-2{\mu }^2}}{\mu }\right)}^2}}$$

Commented by ajfour last updated on 23/Jan/19

$$Itsshorterthantheparametricway,$$$$thanksSir.$$

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