Context: I have been able to prove that the statement below is true (none rigoriously) :
Now I want to find a way to calculate the area of the region bounded by the graph if for all $y$, $x$ only has two or fewer real solutions to $P(x,y)=1$ and for all $x$, $y$ only has two or fewer real solutions to $P(x,y)=1$ or at least knowing how to estimate it. The approach I choose to estimate the bounded region is to find the greatest distance from the origin of a point on the curve $P(x,y)=1$ can be to calculate an area of a circle with that radius. This would serve as the upper bound, doing the same with the shortest distance we can get a lower bound. I tried the simplest case possible, an ellipse."
Question: What is the greatest and shortest distance from the origin to a point on the ellipse?
Follow up question: Is there also a method that generalizes to finding the greatest and shortest distance from the origin to $x^{2n}+j_{n-1}x^{2(n-1)}+$ ... $j_{1}x^2$+ $y^{2n}+k_{n-1}y^{2(n-1)}+$ ... $k_{1}y^2$ = 1?
Well if your ellipse is defined by $$ {x^2\over a^2}+{y^2\over b^2}=1 $$ then the shortest distance is $\min(a,b)$ and the longest distance is $\max(a,b)$.
You can use Lagrange multipliers to find the minimum or maximum of $x^2+y^2$ given that restriction.
For example for the ellipse define $g(x,y)=x^2/a^2+y^2/b^2-1=0$ and $f(x,y)=x^2+y^2$ then we want $\lambda\;\text{grad} g=\text{grad} f$ which gives us $$ \lambda 2x/a^2 = 2x\\ \lambda 2y/b^2 = 2y\\ $$ In the above case either the solutions are $x=0$, $y=0$ or $\lambda/a^2=1$ and $\lambda/b^2=1$, which isn't possible unless $a=b$, so the solutions are at $x=0$ or $y=0$, and you have to explicitly check which is maximum or minimum.