Solve-the-following-equation-simultaneously-and-find-the-stationary-points-2xy-2-c-2-4x-3-y-2-2xy-4-0-1-2x-2-yc-2-2x-4-y-4x-2-y-3-0-2-Please-I-need-a-well-detail-cal

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Question Number 203565 by Mastermind last updated on 22/Jan/24

$$\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equation}\mathrm{simultaneously}$$$$\mathrm{and}\mathrm{find}\mathrm{the}\mathrm{stationary}\mathrm{points}:$$$${2\mathrm{xy}}^2{\mathrm{c}}^2-{4\mathrm{x}}^3{\mathrm{y}}^2-{2\mathrm{xy}}^4=0—–\left(1\right)$$$${2\mathrm{x}}^2{\mathrm{yc}}^2-{2\mathrm{x}}^4\mathrm{y}-{4\mathrm{x}}^2{\mathrm{y}}^3=0—–\left(2\right)$$$$$$$$$$$$\mathrm{Please},\mathrm{I}\mathrm{need}\mathrm{a}\mathrm{well}\mathrm{detail}\mathrm{calculation}$$$$\mathrm{Thank}\mathrm{you}$$

Answered by a.lgnaoui last updated on 22/Jan/24

$$\left(1\right){2\mathrm{xy}}^2({\mathrm{c}}^2-{2\mathrm{x}}^2-{\mathrm{y}}^2)=0$$$$$$$$\left(2\right){2\mathrm{x}}^2\mathrm{y}({\mathrm{c}}^2-{\mathrm{x}}^2-2)=0$$$$\left(3\right)\{\begin{array}{l}{\mathrm{x}}^2={\mathrm{c}}^2-2\\ {\mathrm{y}}^2+{2\mathrm{x}}^2={\mathrm{c}}^2\end{array}{\mathrm{y}}^2={\mathrm{c}}^2-2({\mathrm{c}}^2-2)$$$$\Rightarrow {\mathrm{y}}^2=4-{\mathrm{c}}^2$$$$\{\begin{array}{l}{\mathrm{x}}^2={\mathrm{c}}^2-2\\ {\mathrm{y}}^2=4-{\mathrm{c}}^2\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=2\end{array}$$$$\mathrm{y}=\sqrt{2-{\mathrm{x}}^2}$$$$=\sqrt{4-{\mathrm{c}}^2}2-{\mathrm{x}}^2=4-{\mathrm{c}}^2$$$$\mathrm{x}=\sqrt{{\mathrm{c}}^2-2}$$$$(\mathrm{x},\mathrm{y})=\{(0,0);(\sqrt{{\mathrm{c}}^2-2},\sqrt{4-{\mathrm{c}}^2});$$$$\mathrm{pour}\mathrm{x}=0\mathrm{or}\mathrm{y}=0\mathrm{c}=\sqrt{2}$$$$\mathrm{x}=0\mathrm{and}\mathrm{y}=0\mathrm{c}=0$$$$(\mathrm{x};\mathrm{y},\mathrm{c})=\{(0,0,0);(0,\sqrt{2},\sqrt{2});$$$$(0,\sqrt{4-{\mathrm{c}}^2},\mathrm{c})(\sqrt{{\mathrm{c}}^2-2},0,\mathrm{c})\}$$$$$$$$\left(1\right)-\left(2\right)\Rightarrow ({\mathrm{c}}^2\mathrm{x}-{\mathrm{x}}^3-2\mathrm{x})-{\mathrm{c}}^2\mathrm{y}-$$$$+{2\mathrm{x}}^2\mathrm{y}+{\mathrm{y}}^3=0$$$$({\mathrm{y}}^3-{\mathrm{x}}^3-{\mathrm{c}}^2(\mathrm{y}-\mathrm{x})+2\mathrm{x}(\mathrm{xy}-1)=0$$$$(\mathrm{y}-\mathrm{x})[({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2-{\mathrm{c}}^2)+2\mathrm{x}(\mathrm{xy}-1)$$$$\mathrm{y}=\sqrt{2-{\mathrm{x}}^2}\Rightarrow$$$$$$$$(\sqrt{2-{\mathrm{x}}^2}-\mathrm{x})(2+\mathrm{x}\sqrt{2-{\mathrm{x}}^2}-{\mathrm{c}}^2)+2\mathrm{x}(\mathrm{x}\sqrt{2-{\mathrm{x}}^2}-1)=0$$$$$$$$\Rightarrow 2\sqrt{2-{\mathrm{x}}^2}+\mathrm{x}(2-{\mathrm{x}}^2)-4\mathrm{x}-{\mathrm{c}}^2+$$$${2\mathrm{x}}^2\left(\sqrt{2-{\mathrm{x}}^2}\right)=0$$$$$$$$2\sqrt{2-{\mathrm{x}}^2}({\mathrm{x}}^2+1)-2\mathrm{x}({\mathrm{x}}^2+1)-{\mathrm{c}}^2=0$$$$2({\mathrm{x}}^2+1)(\sqrt{2-{\mathrm{x}}^2}-\mathrm{x})={\mathrm{c}}^2$$$$\sqrt{2-{\mathrm{x}}^2}=\frac{{\mathrm{c}}^2}{2({\mathrm{x}}^2+1)}+\mathrm{x}$$$$2-{\mathrm{x}}^2=\frac{{\mathrm{c}}^4}{4{({\mathrm{x}}^2+1)}^2}+{\mathrm{x}}^2+\frac{{\mathrm{c}}^2\mathrm{x}}{({\mathrm{x}}^2+1)}$$$$8{({\mathrm{x}}^2+1)}^4-{4\mathrm{c}}^2\mathrm{x}({\mathrm{x}}^2+1)-{\mathrm{c}}^4=0$$$$$$$$\mathrm{x}=0\mathrm{c}=\sqrt{2}\Rightarrow (0,0,\sqrt{2})$$$$$$$$$$$$\mathrm{S}\left(\mathrm{Totale}\right)=$$$$\{(0,0,0)(0,0,\sqrt{2})(0,\sqrt{2,}\sqrt{2})(0,\sqrt{4-{\mathrm{c}}^2},0),$$$$(\sqrt{{\mathrm{c}}^2-2},0,\mathrm{c})\}$$$$$$

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