When $px^4+qxy+ry^2+sy+t=0\ (p,q,r,s,t\in\mathbb R)$ represents a simple closed curve on the $xy$ plane, can we represent the area enclosed by this curve by $p,q,r,s,t$? If yes, then how?
Example 1 : $2x^4-2xy+y^2+6y+5=0$ represents a simple closed curve. (see here)
Example 2 : $2x^4-2xy-y^2+6y+5=0$ does not represent a simple closed curve. (see here)
The case $ax^4-by^2\pm\ldots=0$ with $a,b>0$ is an open curve pretty much for the same reason that $ax^2-by^2\pm\ldots=0$ is an open curve as well, called hyperbola. Then all you have to do is compute
$y_{_{1,2}}(x)=\dfrac{-(s+qx)\pm\sqrt{(s+qx)^2-4r(t+px)^4}}{2r}$ , the area being $\displaystyle\int_v^w|y_{_1}(x)-y_{_2}(x)|~dx=$
$=\displaystyle\frac1r\int_v^w\sqrt{(s+qx)^2-4r(t+px)^4}~dx$, where v and w are the left-most and right-most values
of x, i.e., the extrema of $x(y)$, which are to be found amongst the roots of $x'(y)=0$. Granted, this type of integral does not usually possess a closed form expression, and even the exact values of v or w are hard to come by.