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Question Number 64709 by ajfour last updated on 20/Jul/19

Commented by ajfour last updated on 20/Jul/19

$$IfregionsAandBhavethesame$$$$area,andbothparabolashave$$$$thesameshapebuttheiraxis\mathrm{\perp }to$$$$eachother,thendetermine$$$$theequationoftheredparabola.$$

Answered by mr W last updated on 22/Jul/19

$$saybothtouchatpointP(-h,h^2)$$$$y=x^2$$$$\frac{dy}{dx}=2x$$$$eqn.ofredparabola:$$$$x=-s+{(y-t)}^2$$$$\Rightarrow y=t\pm \sqrt{x+s}$$$$\frac{dx}{dy}=2(y-t)\Rightarrow \frac{dy}{dx}=\frac{1}{2(y-t)}$$$$$$$$atpointP:$$$$-h=-s+{(h^2-t)}^2$$$$m=-2h=\frac{1}{2(h^2-t)}$$$$\Rightarrow t=h^2+\frac{1}{4h}$$$$\Rightarrow s=h+\frac{1}{16h^2}$$$$$$$$intersectionofparabolas:$$$$x=-s+{(x^2-t)}^2$$$$\Rightarrow x^4-2{tx}^2-x+t^2-s=0$$$$\Rightarrow x^4-(2h^2+\frac{1}{2h})x^2-x+(h^4-\frac{h}{2})=0$$$${(x+h)}^2(x^2+qx+r)=0$$$$(x^2+2hx+h^2)(x^2+qx+r)=0$$$$x^4+(2h+q)x^3+(h^2+r+2hq)x^2+(h^{2}q+2hr)x+h^{2}r=0$$$$\Rightarrow 2h+q=0\Rightarrow q=-2h$$$$\Rightarrow h^2+r+2hq=-2h^2-\frac{1}{2h}$$$$\Rightarrow h^{2}q+2hr=-1$$$$\Rightarrow h^{2}r=h^4-\frac{h}{2}\Rightarrow r=h^2-\frac{1}{2h}$$$$h^2+r+2hq=h^2+h^2-\frac{1}{2h}-4h^2=-2h^2-\frac{1}{2h}\Rightarrow ok$$$$h^{2}q+2hr=-2h^3+2h(h^2-\frac{1}{2h})=-1\Rightarrow ok$$$$\Rightarrow x^2-2hx+h^2-\frac{1}{2h}=0$$$$\Rightarrow x=h\pm \sqrt{h^2-(h^2-\frac{1}{2h})}=h\pm \frac{1}{\sqrt{2h}}$$$$$$$$A={\int }_{-h}^{h-\frac{1}{\sqrt{2h}}}[t-\sqrt{x+s}-x^2]dx$$$$={[tx-\frac{2}{3}{(x+s)}^{\frac{3}{2}}-\frac{1}{3}{x}^3]}_{-h}^{h-\frac{1}{\sqrt{2h}}}$$$$=(h^2+\frac{1}{4h})(2h-\frac{1}{\sqrt{2h}})-\frac{2}{3}{(h-\frac{1}{\sqrt{2h}}+h+\frac{1}{16h^2})}^{\frac{3}{2}}+\frac{2}{3}{(-h+h+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{1}{3}{(h-\frac{1}{\sqrt{2h}})}^3-\frac{1}{3}{h}^3$$$$=(h^2+\frac{1}{4h})(2h-\frac{1}{\sqrt{2h}})-\frac{2}{3}{(2h-\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}+\frac{2}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{1}{3}{(h-\frac{1}{\sqrt{2h}})}^3-\frac{1}{3}{h}^3$$$$$$$$A+B={\int }_{-h}^{h+\frac{1}{\sqrt{2h}}}[t+\sqrt{x+s}-x^2]dx+{\int }_{-s}^{-h}2\sqrt{x+s}dx$$$$={[tx+\frac{2}{3}{(x+s)}^{\frac{3}{2}}-\frac{1}{3}{x}^3]}_{-h}^{h+\frac{1}{\sqrt{2h}}}+\frac{4}{3}{\left[{(x+s)}^{\frac{3}{2}}\right]}_{-s}^{-h}$$$$=(h^2+\frac{1}{4h})(2h+\frac{1}{\sqrt{2h}})+\frac{2}{3}{(h+\frac{1}{\sqrt{2h}}+h+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{2}{3}{(-h+h+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{1}{3}{(h+\frac{1}{\sqrt{2h}})}^3-\frac{1}{3}{h}^3+\frac{4}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}$$$$=(h^2+\frac{1}{4h})(2h+\frac{1}{\sqrt{2h}})+\frac{2}{3}{(2h+\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{2}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{1}{3}{(h+\frac{1}{\sqrt{2h}})}^3-\frac{1}{3}{h}^3+\frac{4}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}$$$$$$$$A+B=2A$$$$(h^2+\frac{1}{4h})(2h+\frac{1}{\sqrt{2h}})+\frac{2}{3}{(2h+\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{2}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{1}{3}{(h+\frac{1}{\sqrt{2h}})}^3-\frac{1}{3}{h}^3+\frac{4}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}=2(h^2+\frac{1}{4h})(2h-\frac{1}{\sqrt{2h}})-\frac{4}{3}{(2h-\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}+\frac{4}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{2}{3}{(h-\frac{1}{\sqrt{2h}})}^3-\frac{2}{3}{h}^3$$$$\Rightarrow (h^2+\frac{1}{4h})(\frac{3}{\sqrt{2h}}-2h)+\frac{2}{3}{(2h+\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}+\frac{4}{3}{(2h-\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{2}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{1}{3}{(h+\frac{1}{\sqrt{2h}})}^3+\frac{2}{3}{(h-\frac{1}{\sqrt{2h}})}^3+\frac{1}{3}{h}^3=0$$$$\Rightarrow h=2.55781774\Rightarrow h^2=6.542431$$$$eqn.ofredparabola:$$$$\Rightarrow x=-h-\frac{1}{16h^2}+{(y-h^2-\frac{1}{4h})}^2$$$$i.e.x=-2.5673708+{(y-6.6401712)}^2$$$$orx=y^2-13.280342y+41.524503$$

Commented by MJS last updated on 21/Jul/19

$$\mathrm{the}\mathrm{method}\mathrm{is}\mathrm{good}\mathrm{but}\mathrm{I}\mathrm{think}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}$$$$\mathrm{error}\mathrm{somewhere}.\mathrm{the}\mathrm{equation}\mathrm{for}h\mathrm{seems}$$$$\mathrm{to}\mathrm{have}\mathrm{no}\mathrm{solution}$$

Commented by MJS last updated on 21/Jul/19

$$\mathrm{I}\mathrm{tried}\mathrm{it}\mathrm{similar}\mathrm{but}\mathrm{mirrored}\mathrm{at}y=x$$$$\mathrm{parabola}1:y^2=x$$$$\mathrm{parabola}2:y={(x-p)}^2-q$$$$\mathrm{I}\mathrm{put}\mathrm{the}\mathrm{tangential}\mathrm{point}\mathrm{as}\left(\begin{array}{c}\alpha \\ -\sqrt{\alpha }\end{array}\right)$$$$\mathrm{this}\mathrm{leads}\mathrm{to}\sqrt[4]{\alpha }\mathrm{in}\mathrm{some}\mathrm{terms}\mathrm{so}\mathrm{I}\mathrm{put}\alpha ={\beta }^4$$$$\mathrm{I}\mathrm{get}\mathrm{these}\mathrm{equations}:$$$$\mathrm{parabola}1:y=\pm \sqrt{x}$$$$\mathrm{parabola}2:y=x^2-\frac{4{\beta }^6+1}{2{\beta }^2}x+\frac{{\beta }^2(2{\beta }^6-1)}{2}$$$$\mathrm{I}\mathrm{get}\mathrm{these}\mathrm{points}\mathrm{of}\mathrm{interest}:$$$$P_1=\left(\begin{array}{c}{\beta }^4-\sqrt{2}\beta +\frac{1}{2{\beta }^2}\\ {\beta }^2-\frac{\sqrt{2}}{2\beta }\end{array}\right)\left[1^{\mathrm{st}}\mathrm{intersection}\right]$$$$P_2=\left(\begin{array}{c}\frac{4{\beta }^6+1}{4{\beta }^2}-\frac{\sqrt{16{\beta }^6+1}}{4{\beta }^2}\\ 0\end{array}\right)\left[1^{\mathrm{st}}\mathrm{zero}\right]$$$$P_3=\left(\begin{array}{c}{\beta }^4\\ -{\beta }^2\end{array}\right)\left[2^{\mathrm{nd}}\mathrm{intersection}\right]$$$$P_4=\left(\begin{array}{c}\frac{4{\beta }^6+1}{4{\beta }^2}+\frac{\sqrt{16{\beta }^6+1}}{4{\beta }^2}\\ 0\end{array}\right)\left[2^{\mathrm{nd}}\mathrm{zero}\right]$$$$P_5=\left(\begin{array}{c}{\beta }^4+\sqrt{2}\beta +\frac{1}{2{\beta }^2}\\ {\beta }^2+\frac{\sqrt{2}}{2\beta }\end{array}\right)\left[3^{\mathrm{rd}}\mathrm{intersection}\right]$$$$\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{integrals}\mathrm{then}\mathrm{leads}\mathrm{to}$$$${\beta }^{12}-3\sqrt{2}{\beta }^9+\frac{3}{4}{\beta }^6-\frac{3\sqrt{2}}{8}{\beta }^3+\frac{1}{64}=0$$$$[\mathrm{ignoring}\mathrm{an}\mathrm{absolute}\mathrm{value}\Rightarrow \beta ⩾\frac{1}{\sqrt[6]{2}}\approx .890899\Rightarrow \alpha ⩾\frac{1}{\sqrt[3]{4}}\approx .629961]$$$$\beta =\sqrt[3]{t};\beta ⩾\frac{1}{\sqrt[6]{2}}>0\wedge t⩾\frac{1}{\sqrt{2}}>0$$$$t^4-3\sqrt{2}{t}^3+\frac{3}{4}{t}^2-\frac{3\sqrt{2}}{8}t+\frac{1}{64}=0$$$$\mathrm{I}\mathrm{solved}\mathrm{this}\mathrm{with}\mathrm{my}\mathrm{usual}\mathrm{method}\mathrm{putting}$$$$(t-a-\sqrt{b})(t-a+\sqrt{b})(t-c-\sqrt{d})(t-c+\sqrt{d})=0$$$$\mathrm{and}\mathrm{got}$$$$a=-1+\frac{3\sqrt{2}}{4}b=2-\frac{3\sqrt{2}}{2}c=1+\frac{3\sqrt{2}}{4}d=2+\frac{3\sqrt{2}}{2}$$$$t\in \mathbb{R}\wedge t⩾\frac{1}{\sqrt{2}}\Rightarrow t=1+\frac{3\sqrt{2}}{4}+\frac{\sqrt{8+6\sqrt{2}}}{2}$$$$\Rightarrow \beta =\sqrt[3]{t}\Rightarrow \alpha =\sqrt[3]{t^4}$$$$\beta =\sqrt[3]{1+\frac{3\sqrt{2}}{4}+\frac{\sqrt{8+6\sqrt{2}}}{2}}\approx 1.59932$$$$\alpha =\sqrt[3]{\frac{4481}{64}+\frac{99\sqrt{2}}{2}+\frac{3(65+46\sqrt{2})\sqrt{4+3\sqrt{2}}}{8}}\approx 6.54243$$$$\mathrm{parabola}2:y=x^2-13.2803x+41.5245$$

Commented by mr W last updated on 22/Jul/19

$$thankyouforcheckingandforthe$$$$greatsolutionsir!$$$$ihadamistakeinthefinalequation.$$$$nowitshouldbecorrect.oursolutions$$$$arethesamenow.igot$$$$h^2=6.542431$$$$h^{2}is\alpha inyourworking.$$

Commented by MJS last updated on 22/Jul/19

$$(h^2+\frac{1}{4h})(\frac{3}{\sqrt{2h}}-2h)+\frac{2}{3}{(2h+\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}+\frac{4}{3}{(2h-\frac{1}{\sqrt{2h}}+\frac{1}{16h^2})}^{\frac{3}{2}}-\frac{2}{3}{\left(\frac{1}{16h^2}\right)}^{\frac{3}{2}}-\frac{1}{3}{(h+\frac{1}{\sqrt{2h}})}^3+\frac{2}{3}{(h-\frac{1}{\sqrt{2h}})}^3+\frac{1}{3}{h}^3=0$$$$\mathrm{we}\mathrm{know}h>0$$$$2h+\frac{1}{\sqrt{2h}}+\frac{1}{16h^2}={\left(\frac{4\sqrt{2}h\sqrt{h}+1}{4h}\right)}^2$$$$2h-\frac{1}{\sqrt{2h}}+\frac{1}{16h^2}={\left(\frac{4\sqrt{2}h\sqrt{h}-1}{4h}\right)}^2$$$$[\mathrm{believing}4\sqrt{2}h\sqrt{h}-1>0\Rightarrow h>\frac{\sqrt[3]{2}}{4}\approx .314980]$$$$\Rightarrow \mathrm{the}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{transformed}\mathrm{to}$$$$-\frac{64h^6-192\sqrt{2}{h}^{\frac{9}{2}}+48h^3-24\sqrt{2}{h}^{\frac{3}{2}}+1}{48h^3}=0$$$$h^6-3\sqrt{2}{h}^{\frac{9}{2}}+\frac{3}{4}{h}^3-\frac{3\sqrt{2}}{8}{h}^{\frac{3}{2}}+\frac{1}{64}=0$$$$h={\beta }^2[\beta >0]$$$${\beta }^{12}-3\sqrt{2}{\beta }^9+\frac{3}{4}{\beta }^6-\frac{3\sqrt{2}}{8}{\beta }^3+\frac{1}{64}=0$$$$...$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{my}\mathrm{solution}$$

Commented by mr W last updated on 22/Jul/19

$$ajfoursir:$$$$youareabsolutelyright!good$$$$observation!ilostatinypartinthe$$$$areaBindeed.$$$$ihavefixedtheresult.$$

Commented by ajfour last updated on 22/Jul/19

$$ThankyouSir,butMjS{Sir}^{\prime }s$$$$answerisperfect.wewontknow$$$$howtogettheanswerinradical$$$$form,fromoureq.inh.$$

Commented by ajfour last updated on 22/Jul/19

$$Isaluteyousir!$$

Commented by mr W last updated on 22/Jul/19

$$thesolutionofMJSsirisindeeda$$$$perfectone.eventhoughiknewthat$$$$icouldtransfermyfinalequationfor$$$$hintoapolynomialone,butdueto$$$$thehighpoweroftheeqn.igaveup.$$$$becausei{wouldn}^{\prime }tbeabletosolveit.$$

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