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Question Number 102838 by ajfour last updated on 11/Jul/20

Commented by ajfour last updated on 11/Jul/20

$$Ifregions1and2haveequal$$$$areas,findOB=yintermsof$$$$OA=a.$$

Commented by ajfour last updated on 11/Jul/20

$$\frac{x}{a}+\frac{y}{b}=1\Rightarrow bx+ay=ab$$

Answered by mr W last updated on 11/Jul/20

$$OB=b$$$$intersectionatP(p,p^2)$$$$\frac{p}{a}+\frac{p^2}{b}=1$$$$\Rightarrow b=\frac{{ap}^2}{a-p}$$$$\frac{p^3}{3}+\frac{p^2(a-p)}{2}=\frac{1}{2}{\left(\frac{a}{a-p}\right)}^2\frac{p^2(a-p)}{2}$$$$\Rightarrow 2p^2-8ap+3a^2=0$$$$\Rightarrow p=\frac{(4-\sqrt{10})a}{2}$$$$\Rightarrow b=\frac{5\sqrt{10}-14}{6}{a}^2\approx 0.302a^2$$

Commented by mr W last updated on 11/Jul/20

Commented by ajfour last updated on 11/Jul/20

$$Sir,\frac{p^3}{3}+\frac{p^2(a-p)}{2}=\frac{ab}{4}=\frac{a\left({ap}^2\right)}{4(a-p)}$$$$yesitsthesamething,$$$$\frac{p}{3}+\frac{a-p}{2}=\frac{a^2}{4(a-p)}$$$$2(3a-p)(a-p)=3a^2$$$$2p^2-8ap+3a^2=0$$$$p=\left(\frac{8-\sqrt{64-24}}{4}\right)a=\left(\frac{4-\sqrt{10}}{2}\right)a$$$$\frac{b}{a^2}=\left(\frac{26-8\sqrt{10}}{4}\right)\frac{2}{(\sqrt{10}-2)}$$$$\Rightarrow b=\frac{(13-4\sqrt{10})(\sqrt{10}+2)}{6}{a}^2$$$$\mathit{b}=\left(\frac{5\sqrt{10}-14}{6}\right){\mathit{a}}^2.$$$$(Great;thanksSir,noerror!)$$

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