From ∫036f(36tx)dt=4αf(x), substitute u=36tx to get x36∫0xf(u)du=4αf(x).
Differentiating: f(x)=9α[f(x)+xf′(x)], which simplifies to (9−α)f(x)=αxf′(x).
This gives f(x)f′(x)=αx9−α, integrating to f(x)=Kxα9−α.
For f to be a standard parabola: α9−α=2⇒α=3.
Thus f(x)=Kx2. Using point (2,1): K=41, so f(x)=4x2.
At (−4,β): β=416=4.
Therefore βα=43=64