Let a differentiable function f satisfy the equation ∫_0^36 f((t x)/36) d t=4 α f(x). If y=f(x) is a standard parabola passing through the points…

Let a differentiable function $\displaystyle f$ satisfy the equation $\displaystyle \int_0^{36} f\left(\frac{t x}{36}\right) d t=4 \alpha f(x)$. If $\displaystyle y=f(x)$ is a standard parabola passing through the points $\displaystyle (2,1)$ and $\displaystyle (-4, \beta)$, then $\displaystyle \beta^\alpha$ is equal to $\displaystyle \_\_\_\_$.