Question Number 47425 by ajfour last updated on 09/Nov/18
Commented by ajfour last updated on 09/Nov/18
$${Find}\:{b}\:{if}\:{a}=\mathrm{1}\:{and}\:\bigtriangleup{ABC}\:{has} \\ $$$${minimum}\:{area}. \\ $$
Answered by mr W last updated on 09/Nov/18
![eqn. of ellipse: (x^2 /a^2 )+(y^2 /b^2 )=1 let tan α=(b/a) eqn. of BA: y=(b/a)x+b eqn. of BC: y=−(a/b)x A(a,y_A ),C(a,y_C ) y_A =(b/a)a+b=2b y_C =−(a/b)a=−(a^2 /b) AC=y_A −y_C =2b+(a^2 /b)=((a^2 +2b^2 )/b) S_(ABC) =((AC^2 sin α cos α)/2)=(((a^2 +2b^2 )^2 ab)/(2b^2 (a^2 +b^2 ))) with λ=(b/a) S_(ABC) =(((1+2λ^2 )^2 a^2 )/(2λ(1+λ^2 ))) (dS/dλ)=(a^2 /2)[((8(1+2λ^2 ))/((1+λ^2 )))−(((1+2λ^2 )^2 (1+3λ^2 ))/(λ^2 (1+λ^2 )^2 ))]=0 8λ^2 (1+λ^2 )−(1+2λ^2 )(1+3λ^2 )=0 2λ^4 +3λ^2 −1=0 λ^2 =(((√(17))−3)/4) ⇒λ=(b/a)=((√((√(17))−3))/2)≈0.5299 ⇒b≈0.5299](https://www.tinkutara.com/question/Q47434.png)
$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${let}\:\mathrm{tan}\:\alpha=\frac{{b}}{{a}} \\ $$$${eqn}.\:{of}\:{BA}: \\ $$$${y}=\frac{{b}}{{a}}{x}+{b} \\ $$$${eqn}.\:{of}\:{BC}: \\ $$$${y}=−\frac{{a}}{{b}}{x} \\ $$$${A}\left({a},{y}_{{A}} \right),{C}\left({a},{y}_{{C}} \right) \\ $$$${y}_{{A}} =\frac{{b}}{{a}}{a}+{b}=\mathrm{2}{b} \\ $$$${y}_{{C}} =−\frac{{a}}{{b}}{a}=−\frac{{a}^{\mathrm{2}} }{{b}} \\ $$$${AC}={y}_{{A}} −{y}_{{C}} =\mathrm{2}{b}+\frac{{a}^{\mathrm{2}} }{{b}}=\frac{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} }{{b}} \\ $$$${S}_{{ABC}} =\frac{{AC}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}}=\frac{\left({a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} {ab}}{\mathrm{2}{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${with}\:\lambda=\frac{{b}}{{a}} \\ $$$${S}_{{ABC}} =\frac{\left(\mathrm{1}+\mathrm{2}\lambda^{\mathrm{2}} \right)^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{2}\lambda\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)} \\ $$$$\frac{{dS}}{{d}\lambda}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\frac{\mathrm{8}\left(\mathrm{1}+\mathrm{2}\lambda^{\mathrm{2}} \right)}{\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}−\frac{\left(\mathrm{1}+\mathrm{2}\lambda^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}\lambda^{\mathrm{2}} \right)}{\lambda^{\mathrm{2}} \left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }\right]=\mathrm{0} \\ $$$$\mathrm{8}\lambda^{\mathrm{2}} \left(\mathrm{1}+\lambda^{\mathrm{2}} \right)−\left(\mathrm{1}+\mathrm{2}\lambda^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{3}\lambda^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{2}\lambda^{\mathrm{4}} +\mathrm{3}\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} =\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\lambda=\frac{{b}}{{a}}=\frac{\sqrt{\sqrt{\mathrm{17}}−\mathrm{3}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{5299} \\ $$$$\Rightarrow{b}\approx\mathrm{0}.\mathrm{5299} \\ $$
Commented by ajfour last updated on 10/Nov/18
$${Very}\:{Nice},\:{Sir}! \\ $$