if $(x_0,y_0)$ is local extrema in $ax^2 + by^2 + cxy + dx + ey + l$ then its global too. An exercise on a book asks me to prove that if a point $(x_0,y_0)$ is a local extrema for the function $$f(x,y) = ax^2 + by^2 + cxy + dx + ey + l$$
then it's also a global extrema.
The exercise asks me to consider that $g(t) = f(x_0+ht,y_0+kt)$ is a parabola, but I couldn't understand how it helps. Maybe since it's a parabola, I know that this derivative has to be $0$ in some point, and also that this function $g(t)$ has a global extrema point, but I couldn't derive anything from this. I also tried to find the derivatives, equal them to $0$ and solve the system. I got:
$$x_0 = -\frac{2 b d - c e}{4 a b - c^2}, y_0 = -\frac{
  c d - 2 a e}{-4 a b + c^2}$$
But when I try to find 
$$f\left(-\frac{2 b d - c e}{4 a b - c^2},-\frac{
  c d - 2 a e}{-4 a b + c^2}\right)$$ so I can try to prove:
$$f(x,y)-f(x_0,y_0)$$
is either greater or less $0$ depending on the case (so I can do $f(x,y)
\le f(x_0,y_0)$ and $f(x,y)
\ge f(x_0,y_0)$)
things get hairy.
 A: Lets assume that $(x_0,y_0)$ is a local but not a global minimum. Then there exists some point $(x_1,y_1)$ such that $f(x_1,y_1)<f(x_0,y_0)$ and an $\epsilon>0$ such that $f(x,y)\geq f(x_0,y_0)$ for all $(x,y)$ satisfying $\sqrt{(x-x_0)^2+(y-y_0)^2}<\epsilon$.
Consider the function $g(t):=f((x_0,y_0)+t(x_1-x_0,y_1-y_0))$; this function is just the restriction of $f$ to the line passing through $(x_0,y_0)$ and $(x_1,y_1)$. $g$ is a quadratic function in one real variable, hence we know that it possesses exactly one local minimum or local maximum, which then automatically is global. In our case by construction $g$ has a local minimum at $t=0$. But on the other hand we have $g(1)<g(0)$, a contradiction.
A: Why does the hint provided helps?
Suppose that the local extrema at $(x_0,y_0)$ is a local minimum. If not change $f$ into $-f$.
In order to prove that $(x_0,y_0)$ is a global minimum, you have to prove that for all $(x,y) \in \mathbb R^ 2$:
$$f(x,y) \ge f(x_0,y_0).$$
For $(x,y) \in \mathbb R^ 2 \setminus \{(x_0,y_0)\}$, take $$(h,k)=(x-x_0,y-y_0) \neq (0,0).$$
The map $$g(t)=f(x_0+ht,y_0+kt)$$ is differentiable and represents a parabola. You can see it by substituting $(x_0+ht,y_0+kt)$ in the definition of $f$. The parabola has a minimum for $t = 0$ as $$g^\prime(t)=0 \Longleftrightarrow t=0.$$ Therefore $f(x,y) =g(1) \ge g(0)=f(x_0,y_0)$.
