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Question Number 103986 by ajfour last updated on 18/Jul/20

Commented by ajfour last updated on 18/Jul/20

$$Ifbothcolouredregionshaveequal$$$$area,find{x}_A,x_B.$$

Answered by mr W last updated on 19/Jul/20

$$B(-q,q^2)$$$$A(p,p^2)$$$$-\frac{1}{\tan \theta }=-2q\Rightarrow \tan \theta =\frac{1}{2q}$$$$eqn.ofBA:$$$$y=\frac{1}{2q}(x+q)+q^2$$$$p^2=\frac{1}{2q}(p+q)+q^2$$$$2{qp}^2-p-q(2q^2+1)=0$$$$\Rightarrow p=q+\frac{1}{2q}$$$$-\frac{1}{\tan \phi }=2p=2q+\frac{1}{q}=\frac{2q^2+1}{q}$$$$\Rightarrow \tan \phi =-\frac{q}{2q^2+1}$$$$eqn.ofAC:$$$$y=-\frac{q}{2q^2+1}(x-p)+p^2$$$$y=-\frac{q}{2q^2+1}(x-q-\frac{1}{2q})+{(q+\frac{1}{2q})}^2$$$$areaunderAC=A_1$$$$areaunderBA=A_2$$$$x^2+\frac{q}{2q^2+1}x-(\frac{q}{2q^2+1}+q+\frac{1}{2q})(q+\frac{1}{2q})=0$$$$\mathrm{\Delta }={\left(\frac{q}{2q^2+1}\right)}^2+4(\frac{q}{2q^2+1}+q+\frac{1}{2q})(q+\frac{1}{2q})$$$$={(\frac{q}{2q^2+1}+2q+\frac{1}{q})}^2$$$$6A_1={\left(\mathrm{\Delta }\right)}^{\frac{3}{2}}={(\frac{q}{2q^2+1}+2q+\frac{1}{q})}^3$$$$$$$$x^2-\frac{1}{2q}x-(\frac{1}{2}+q^2)=0$$$$\mathrm{\Delta }=\frac{1}{4q^2}+2+4q^2={(\frac{1}{2q}+2q)}^2$$$$6A_2={\left(\mathrm{\Delta }\right)}^{\frac{3}{2}}={(\frac{1}{2q}+2q)}^3$$$$A_1=2A_2$$$$\Rightarrow \frac{q}{2q^2+1}+2q+\frac{1}{q}=\sqrt[3]{2}(\frac{1}{2q}+2q)$$$$\Rightarrow 8(\sqrt[3]{2}-1)q^4-2(5-3\sqrt[3]{2})q^2-(2-\sqrt[3]{2})=0$$$$\Rightarrow q=\sqrt{\frac{2-\sqrt[3]{2}}{2(\sqrt[3]{2}-1)}}\approx 1.193173$$

Commented by mr W last updated on 18/Jul/20

Commented by ajfour last updated on 19/Jul/20

$$FabulousSir,letmesometime$$$$comprehendingitentirely..$$

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