Question-53492

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

Commented by ajfour last updated on 22/Jan/19

Find parameters a and b of an ellipse that circumscribes an equilateral triangle of side t and even a square of side s.

$${Find}\:{parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{of}\:{an} \\ $$$${ellipse}\:{that}\:{circumscribes}\:{an} \\ $$$${equilateral}\:{triangle}\:{of}\:{side}\:\boldsymbol{{t}}\:{and} \\ $$$${even}\:{a}\:{square}\:{of}\:{side}\:\boldsymbol{{s}}. \\ $$

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

(x^2 /a^2 )+(y^2 /b^2 )=1 (s^2 /(4a^2 ))+(s^2 /(4b^2 ))=1 ⇒(1/a^2 )+(1/b^2 )=(4/s^2 ) (t^2 /(4a^2 ))+((((((√3)t)/2)−b)^2 )/b^2 )=1 ⇒(t^2 /a^2 )+((((√3)t−2b)^2 )/b^2 )=4 ⇒((4t^2 )/s^2 )−(t^2 /b^2 )+((√3)(t/b)−2)^2 =4 with μ=(t/s), λ=(t/b) ⇒4μ^2 −λ^2 +((√3)λ−2)^2 =4 ⇒λ^2 −2(√3)λ+2μ^2 =0 ⇒λ=(t/b)=(√3)+(√(3−2μ^2 )) ⇒(s/b)×μ=(√3)+(√(3−2μ^2 )) ⇒b=(t/( (√3)+(√(3−2μ^2 )))) ⇒a=(s/( (√(4−((s/b))^2 ))))=(s/( (√(4−((((√3)+(√(3−2μ^2 )))/μ))^2 ))))

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

Commented by ajfour last updated on 23/Jan/19

$${Its}\:{shorter}\:{than}\:{the}\:{parametric}\:{way}, \\ $$$${thanks}\:{Sir}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply