I know the condition for quadratic polynomial to be an ellipse-like closed convex curve is
$$ b^2-4ac < 0, $$
where $f(x,y) = ax^2+bxy+cy^2+dx+ey+f = 0.$
Then what's the same condition for cubic polynomial such as following?
\begin{equation} f(x,y) = a_1x^3+a_2x^2y+a_3xy^2+a_4y^3+ a_5x^2+\\a_6xy+a_7y^2+a_8x+a_9y +a_{10} = 0 \end{equation}
I was guessing the condition might be that determinant of $f''(x,y)$ matrix be less than 0, but as element of $f''(x,y)$ are involving terms of $x$ and $y$, I don't think my guess is correct. Any hint would be appreciated.
Added some background below for your convenience. I was estimating egg-shape curve with cubic polynomial. And sometimes, the estimation is good. And in some cases, it's bad although the two datasets are similar. This led to my question.
UPDATE Thanks to people's effort on this question, I was able to clear my mind and decided to impose conditions from the original model (distorted quadratic curve if not rotated) I was trying to estimate. Now, I am using constrained cubic polynomial.
Suppose that $f(x,y)$ has an $x^3$ term. Let $y=y_L$ be large. Then the equation $f(x,y_L)=0$ has a solution, so there are points on the curve arbitrarily far from the origin. The same is true if $f(x,y)$ has a $y^3$ term.
Now suppose there is no $x^3$ or $y^3$ term. Then if our polynomial has degree $3$, there must be an $x^2y$ term or an $xy^2$ term or both. So the cubic has shape $ax^2y+bxy^2+\text{lower order terms}$, where at least one of $a$ or $b$ is non-zero. Let $y=x+t$. Then (unless $b=-a$) our cubic gets transformed to a cubic in $x$ and $t$ with an $x^3$ term. So there are solutions with arbitrarily large $t$, and hence with $(x,y)$ arbitrarily far from the origin. The case $b=-a$ is dealt with by using $y=-x+t$.