How can I plot the set $M:=\left\{\left(\begin{array}{c}x\\y\end{array}\right)\in \mathbb R^2:9x^4-16x^2y^2+9y^4\leq9\right\}$ I need to plot the following set:
$$M:=\left\{\left(\begin{array}{c}x\\y\end{array}\right)\in \mathbb R^2:9x^4-16x^2y^2+9y^4\leq9\right\}$$
I have solved the equation $9x^4-16x^2y^2+9y^4-9=0$ for $y$, thinking this would make the task easier, but it turned out not to bring much help. However this is what I got:
$$y_{1,2,3,4}=\pm \sqrt{\frac89x^2\pm\frac19\sqrt{81-17x^4}}$$
Not so obvious how one can plot any of these functions. I might not see some trick that would help me to visualize the set (maybe switch to polar coordinates? Tried this, but didn't help me either). Of course after having struggled an hour or so, I asked Mathematica for help and got this:

The question is how it can be done without using the magic of Mathematica.
 A: Note that in polar co-ordinates $x=r\cos \theta, y=r\sin \theta$ the function becomes $$r^4(9\cos^4\theta-16\cos^2\theta\sin^2\theta+9\sin^4\theta)$$
Then $\cos^4\theta+\sin^4\theta =(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta$ so the trigonometric part becomes $$9-34\cos^2\theta\sin^2\theta=9-\frac {17}2\sin^2 2\theta$$
Then $\sin^2 2\theta=\frac 12-\frac 12\cos 4\theta$ so we get $$9-\frac {17}4+\frac {17}4\cos4\theta=\frac {19+17\cos4\theta}{4}$$
The boundary of the region is therefore $$r^4=\frac {36}{19+17\cos4\theta}$$
The right-hand side is always positive, and has a maximum value $18$ when $\cos 4\theta=-1$, minimum value $1$ when $\cos 4\theta=1$.
A: In order to plot $(x,y)$, $x$ and $y$ must be real.
Thus condition (1) is:
$$0\le \sqrt{81-17x^4} \implies 0\le |x| \le (81/17)^{1/4}=1.45:=x_{max}$$
The condition (2) is:
$$\sqrt{\frac89x^2-\frac19\sqrt{81-17x^4}}\ge 0 \implies (8x^2)^2 \ge(\sqrt{81-17x^4})^2 \implies |x|\ge 1=:x_{min}$$
(A) So the top portion of the right curve is given by
$$y_{1}(x)=+ \sqrt{\frac89x^2-\frac19\sqrt{81-17x^4}}  \text {  } (x_{min}\le x\le x_{max})$$
(B) the top portion of the left curve is given by
$$y_{1}(x)=+ \sqrt{\frac89x^2-\frac19\sqrt{81-17x^4}}  \text {  } (-x_{max}\le x\le -x_{min})$$
(C) So the bottom portion of the right curve is given by
$$y_{1}(x)=- \sqrt{\frac89x^2-\frac19\sqrt{81-17x^4}}  \text {  } (x_{min}\le x\le x_{max})$$
(D) the bottom portion of the left curve is given by
$$y_{1}(x)=- \sqrt{\frac89x^2-\frac19\sqrt{81-17x^4}}  \text {  } (-x_{max}\le x\le -x_{min})$$
Because (x,y) are symmetric in the original inequality, the top curve and bottom curve are given by swapping x with y in the formulas above.
-mike
A: LLet $X=x^2$ and $Y=y^2$.  The equation becomes the equation of an ellipse.  You can solve that, and should get something like $X=a\cos\theta+b\sin\theta, Y=a\cos\theta-b\sin\theta$.  Then to plot the function in $x,y$ coordinates, $(x,y)=(\pm\sqrt{a\cos\theta+b\sin\theta},\pm\sqrt{a\cos\theta-b\sin\theta})$
