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Question Number 40063 by ajfour last updated on 15/Jul/18

Answered by ajfour last updated on 16/Jul/18

$$Q\equiv (a,\sqrt{R^2-a^2})$$$$TL=QL$$$$\Rightarrow T(-a,\sqrt{R^2-a^2})$$$$eq.oftangent:$$$$-ax+y\sqrt{R^2-a^2}=R^2$$$$x_M=-\frac{R^2}{a}$$$$SM=-R-(-\frac{R^2}{a})=R(\frac{R}{a}-1)$$$$\frac{y_L}{R}=\frac{\sqrt{R^2-a^2}}{a+R}$$$$\Rightarrow y_L=\frac{R\sqrt{R^2-a^2}}{a+R}$$$${QL}^2=a^2+{(\sqrt{R^2-a^2}-\frac{R\sqrt{R^2-a^2}}{a+R})}^2$$$$=a^2+\frac{a^2(R^2-a^2)}{{(a+R)}^2}$$$$=\frac{a^2(2R^2+2aR)}{{(a+R)}^2}=\frac{2a^{2}R}{a+R}$$$${SM}^2={QL}^2$$$$\Rightarrow R^2{(\frac{R}{a}-1)}^2=\frac{2a^{2}R}{a+R}$$$$\Rightarrow R{(R-a)}^2(a+R)=2a^4$$$$\Rightarrow let\frac{a}{R}=z$$$$\Rightarrow {(1-z)}^2(z+1)=2z^4$$$$or2z^4-(z-1)(z^2-1)=0$$$$2z^4-z^3+z^2+z-1=0$$$$\Rightarrow \mathit{z}\approx 0.59887$$$$\mathit{a}\approx 0.59887R$$$$as(a,b)liesonthetangent,$$$$\Rightarrow -a^2+b\sqrt{R^2-a^2}=R^2$$$$\Rightarrow \mathit{b}=\frac{{\mathit{R}}^2+{\mathit{a}}^2}{\sqrt{{\mathit{R}}^2-{\mathit{a}}^2}}$$$$\mathit{b}=\mathit{R}\left(\frac{1+{\mathit{z}}^2}{\sqrt{1-{\mathit{z}}^2}}\right)\approx 1.696512\mathit{R}.$$

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