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Question Number 47425 by ajfour last updated on 09/Nov/18

Commented by ajfour last updated on 09/Nov/18

$$Findbifa=1and△ABChas$$$$minimumarea.$$

Answered by mr W last updated on 09/Nov/18

$$eqn.ofellipse:$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$let\tan \alpha =\frac{b}{a}$$$$eqn.ofBA:$$$$y=\frac{b}{a}x+b$$$$eqn.ofBC:$$$$y=-\frac{a}{b}x$$$$A(a,y_A),C(a,y_C)$$$$y_A=\frac{b}{a}a+b=2b$$$$y_C=-\frac{a}{b}a=-\frac{a^2}{b}$$$$AC=y_A-y_C=2b+\frac{a^2}{b}=\frac{a^2+2b^2}{b}$$$$S_{ABC}=\frac{{AC}^2\sin \alpha \cos \alpha }{2}=\frac{{(a^2+2b^2)}^{2}ab}{2b^2(a^2+b^2)}$$$$with\lambda =\frac{b}{a}$$$$S_{ABC}=\frac{{(1+2{\lambda }^2)}^2{a}^2}{2\lambda (1+{\lambda }^2)}$$$$\frac{dS}{d\lambda }=\frac{a^2}{2}[\frac{8(1+2{\lambda }^2)}{(1+{\lambda }^2)}-\frac{{(1+2{\lambda }^2)}^2(1+3{\lambda }^2)}{{\lambda }^2{(1+{\lambda }^2)}^2}]=0$$$$8{\lambda }^2(1+{\lambda }^2)-(1+2{\lambda }^2)(1+3{\lambda }^2)=0$$$$2{\lambda }^4+3{\lambda }^2-1=0$$$${\lambda }^2=\frac{\sqrt{17}-3}{4}$$$$\Rightarrow \lambda =\frac{b}{a}=\frac{\sqrt{\sqrt{17}-3}}{2}\approx 0.5299$$$$\Rightarrow b\approx 0.5299$$

Commented by ajfour last updated on 10/Nov/18

$$VeryNice,Sir!$$

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