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When plotting following implicit function $x^4 + y^4 = 2xy$, what is the best way to parametrize it? I tried $y=xt$, but then I get $x=\sqrt{2t\over {1+t^4}}$ and $y=\sqrt{2t^3\over {1+t^4}}$, which means I am restricted by $x>0, y>0, t>0$. It is easy to prove that the curve is symmetric about the origin, but still I can't say anything about the second a fourth quadrant...


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2 Answers 2

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That relation requires $xy \ge 0$, which translates to $$ \left [ \begin{array}{c} \left \{ \begin{array}{c} x \ge 0 \\ y \ge 0 \end{array} \right . \\ \left \{ \begin{array}{c} x < 0 \\ y < 0 \end{array} \right . \end{array} \right . $$ So, the curve defined by the relation doesn't go through second and fourth quadrants. Simple $x \to -x$ and $y \to -y$ substitutions show that curve is symmetric w.r.t. origin. So, indeed you may consider only first quadrant by the parametrization you suggested and simply reflect it from the origin due to the symmetry. Otherwise, if you want rigorous derivation, you need to keep in mind that $x^2 = a$ means that $x = \pm \sqrt a$, for $a \ge 0$, not just $\sqrt a$. $$ \left [ \begin{array}{l} \left \{ \begin{array}{c} x = \sqrt{\frac {2t}{1+t^4}} \\ y = \sqrt{\frac {2t^3}{1+t^4}} \end{array} \right . \\ \left \{ \begin{array}{c} x = -\sqrt{\frac {2t}{1+t^4}} \\ y = -\sqrt{\frac {2t^3}{1+t^4}} \end{array} \right . \end{array} \right . $$ and $t \ge 0$.



You can also use angle-based parametrization per @user86418' post $$ x = \pm \sqrt{r \cos \theta} \\ y = \pm \sqrt{r \sin \theta} $$ so curve equation becomes $$ r = 2\sqrt{\sin \theta \cos \theta} = \sqrt{2 \sin 2\theta} $$ By substituting to original equation one can find that $$ \begin{array}{c} x = \pm \sqrt[4]{2\sin 2\theta \cos^2 \theta} \\ y = \pm \sqrt[4]{2\sin 2\theta \sin^2 \theta} \end{array}, \qquad \theta \in \left [ 0, \frac \pi 2\right ] $$

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"Best" may be in the eye of the beholder, but your curve is parametrized in polar coordinates as $$ r = \sqrt{\frac{\sin(2\theta)}{\cos^4\theta + \sin^4\theta}} = \sqrt{\frac{\sin(2\theta)}{1 - \frac{1}{2} \sin^2(2\theta)}},\qquad 0 \leq \theta \leq \pi/2,\quad \pi \leq \theta \leq 3\pi/2. $$

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You don't have to use traditional polar coordinates. I updated my answer. –  Kaster Mar 14 at 23:11

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