I wonder if such a task is possible: we have a curve defined implicitly with:
$x^4+y^3-xy-1=0\qquad(1)$
I want to find a parabola in a general form of:
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
... which approximates the curve the best at the point $(1,1)$.
My idea was to treat the equation $(1)$ as a function $F(x,y)$ and expand it into a Taylor polynomial $W(x,y)$ of a degree 2:
$W(x,y)=F(1,1)+df(1,1)(x-1,y-1)+\frac{1}{2}d^2f(1,1)(x-1,y-1)$
... where:
$d^nf(1,1)(x-x_0,y-y_0)=\sum\limits_{i=0}^{n}{n\choose i}\frac{\partial^nf}{\partial x^{n-i}\partial y^i}(x_0,y_0)\cdot(x-x_0)^{n-i}(y-y_0)^i$
In the end I got the following formula for the "parabola":
$3(x-1)+2(y-1)+12(x-1)^2-2(x-1)(y-1)+6(y-1)^2=\\=6x^2-xy+3y^2-8x-3y+3$
However to my surprise it turned out to be an ... ellipse instead:
The question is, is it the only valid answer (if it is correct at all)? Can I "explode" this ellipse into a parabola that would satisfy my needs? And since we're at it, a side-question: what about transforming it into a hyperbola, is it possible?
Cheers
(edited 5.6.2014): If $F(x,y)=0$ represents a curve, then it is not true that its second order Taylor approximation in $(a,b)$ is $dF(a,b)(x-a, y-b)+d^2 F(a,b)(x-a, y-b)=0$. (Note, that if $d^2 F$ is positive definite, it will always be an ellipse.)
You could compute derivatives of the implicite function $y(x)$ in a neighborhood of $(1,1)$. Differentiate your original equation and you get $4x^3+3y^2 y' - y - xy'=0$, which implies $y'(1)=-3/2$. Similarly, you compute $y''(1)$ and your parabole is then just $y=y(1)+(x-1) y'(1) + \frac{y''(1)}{2} (x-1)^2$.
However, if you are not interested in properties of the function $y(x)$ but rather in some geometric properties of the zero set itself, then you should specify more exactly what you mean by "best approximation" and why should such thing be a parabola at all. One approximation possibility could be the osculating circle, which is (among other properties) invariant wrt. Euclidean transformations of your curve.