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Question Number 216655 by abdelsalam last updated on 14/Feb/25

Answered by issac last updated on 14/Feb/25

$$y^4=y^2-x^2$$$$y^4-y^2+\frac{1}{4}-\frac{1}{4}=-x^2$$$${(y^2-\frac{1}{2})}^2=\frac{1}{4}-x^2$$$$y^2=\frac{1}{2}\pm \sqrt{\frac{1}{4}-x^2}$$$$y=\pm \sqrt{\frac{1}{2}\pm \sqrt{\frac{1}{4}-x^2}}$$$$\frac{\mathrm{d}}{\mathrm{d}x}{y}^4=\frac{\mathrm{d}}{\mathrm{d}x}(y^2-x^2)$$$$4y^3\frac{\mathrm{d}y}{\mathrm{d}x}=2y\frac{\mathrm{d}y}{\mathrm{d}x}-2x$$$$(4y^3-2y)\frac{\mathrm{d}y}{\mathrm{d}x}=-2x$$$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{2y^3-y}=0$$$$2y^3-y\ne 0,x=0\therefore x=0$$$$\therefore \mathrm{that}\mathrm{Curve}\mathrm{haves}\mathrm{two}\mathrm{Horizental}\mathrm{slope}$$

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