Minimum value of a general two degree curve

Question

Let F(x, y) = $ax^2 + by^2 + 2gx + 2fy + 2hxy + c$ To find its minimum value I use the following logic abd procedure : When you put a point (x, y) in equation of a conic it gives a positive value when point is outside the conic and negative when inside. So the least value should be at the centre of the conic. I partially differentiate the equation wrt x and then wrt y and solve the linear equations obtained to find the centre. Putting these x and y in equation I get minimum value. My question : I understand that I will get minimum value when the equation represents a conic with centre such as ellipse but will this procedure work for equations representing other conics (eg parabola which doesn't have a centre) if yes then why?

Answer 1

Be careful when you assume that "the least value should be at the centre of the conic", in general there is no guarantee that this is true. In the case of ellipses, it works for symmetry reasons.

But by some irony, your method to find the center is the right method to find the extrema for general bivariate functions !

Indeed, a local minimum is such that the partial derivatives vanish (the tangent plane is horizontal). This is called a stationary point. Anyway, a stationary point can be a minimum, a maximum or a saddle point, and you also need to check the second derivatives.

For your conic, the stationary points are located by

$$\begin{cases}ax+hy+g=0,\\hx+by+f=0\end{cases}$$ in all cases.

The second derivatives are constant and form the matrix

$$\begin{pmatrix}a&h\\h&b\end{pmatrix}.$$

If this matrix is positive definite $(ab-h^2>0,a>0$), you have a minimum.

If this matrix is negative definite $(ab-h^2>0,a<0$), you have a maximum.

If this matrix is semi-positive/negative definite $(ab-h^2=0$), you have no stationary point.

Otherwise ($ab-h^2<0$) you have a saddle point.