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Question Number 122399 by ajfour last updated on 16/Nov/20

Commented by ajfour last updated on 16/Nov/20

$$Findintermsofradiiaandb$$$$theminimumlengthofcommon$$$$tangentEFbetweentheaxes.$$

Answered by mr W last updated on 16/Nov/20

$$L=\frac{a}{\sin \alpha }+\frac{b}{\tan \beta }+\sqrt{{(b+a)}^2-{(b-a)}^2}$$$$=\frac{a}{\tan \alpha }+\frac{b}{\tan (\frac{\pi }{4}-\alpha )}+2\sqrt{ab}$$$$=\frac{a}{\tan \alpha }+\frac{b(1+\tan \alpha )}{1-\tan \alpha }+2\sqrt{ab}$$$$L=\frac{a}{\mathrm{t}}+\frac{2b}{1-t}-b+2\sqrt{ab}$$$$\frac{dL}{dt}=-\frac{a}{t^2}+\frac{2b}{{(1-t)}^2}=0$$$${(1-\frac{1}{t})}^2=\frac{2b}{a}$$$$\Rightarrow t=\tan \alpha =\frac{1}{\sqrt{\frac{2b}{a}}+1}$$$$L_{min}=a(\sqrt{\frac{2b}{a}}+1)+2b(\frac{1}{\sqrt{\frac{2b}{a}}}+1)-b+2\sqrt{ab}$$$$=a+b+2(1+\sqrt{2})\sqrt{ab}$$

Commented by ajfour last updated on 16/Nov/20

$$thanksforsolvingsir!$$

Answered by ajfour last updated on 16/Nov/20

$$letL=x+y+2\sqrt{ab}$$$${\tan }^{-1}\frac{b}{x}+{\tan }^{-1}\frac{a}{y}=\frac{\pi }{4}$$$$\frac{\frac{b}{x}+\frac{a}{y}}{1-\frac{ab}{xy}}=1\Rightarrow ax+by=xy-ab$$$$L=x+a\left(\frac{x+b}{x-b}\right)+2\sqrt{ab}$$$$=x+a+\frac{2ab}{x-b}+2\sqrt{ab}$$$$\frac{dL}{dx}=1-\frac{2ab}{{(x-b)}^2}=0$$$$\Rightarrow x-b=\sqrt{2ab}$$$$\Rightarrow x+a=a+b+\sqrt{2ab}$$$$L_{min}=a+b+2\sqrt{2ab}+2\sqrt{ab}$$$$L_{min}=a+b+2(\sqrt{2}+1)\sqrt{ab}$$

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