Prove that a function is the quadratic form associated to

Homework Statement

Let G:R2$\rightarrow$R be a C2 function such that G(tx,ty)=t2G(x, y). Show that:

2G(x,y)=(x,y).HG(0,0).(x,y)t

The Attempt at a Solution

G is C2, so its Taylor expansion is:

G(x,y) = G(0,0) + $\nabla$G(0,0).(x,y) + $\frac{1}{2}$(x,y).HG(c).(x,y)t,

where c lies on the line segment that goes from (0,0) to (x,y).

Using that G(tx,ty)=t2G(x,y) I get that G(0,0) and the linear term equals 0.

The problem is that I have HG(c) in the quadratic term, but I need HG(0,0).

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