Suppose $a$ and $b$ are two variables that satisfy $\textstyle\int_0^2(-ax^2+b)dx=0$. What is $\tfrac{a}{b}$?
1. We start with the given integral equation:
   \[
   \int_0^2 (-ax^2 + b) \, dx = 0
   \]

2. To solve this, we first find the indefinite integral of the integrand \(-ax^2 + b\):
   \[
   \int (-ax^2 + b) \, dx = -\frac{a}{3} x^3 + bx + C
   \]
   where \(C\) is the constant of integration.

3. Let \(F(x) = \int (-ax^2 + b) \, dx = -\frac{a}{3} x^3 + bx + C\). We need to evaluate this definite integral from 0 to 2:
   \[
   \int_0^2 (-ax^2 + b) \, dx = F(2) - F(0)
   \]

4. Calculate \(F(2)\):
   \[
   F(2) = -\frac{a}{3} (2)^3 + b(2) + C = -\frac{8a}{3} + 2b + C
   \]

5. Calculate \(F(0)\):
   \[
   F(0) = -\frac{a}{3} (0)^3 + b(0) + C = C
   \]

6. Substitute \(F(2)\) and \(F(0)\) back into the definite integral:
   \[
   \int_0^2 (-ax^2 + b) \, dx = \left( -\frac{8a}{3} + 2b + C \right) - C = -\frac{8a}{3} + 2b
   \]

7. Set the integral equal to zero as given in the problem:
   \[
   -\frac{8a}{3} + 2b = 0
   \]

8. Solve for \(a\) in terms of \(b\):
   \[
   -\frac{8a}{3} + 2b = 0 \implies -\frac{8a}{3} = -2b \implies \frac{8a}{3} = 2b \implies 8a = 6b \implies a = \frac{6b}{8} = \frac{3b}{4}
   \]

9. Therefore, the ratio \(\frac{a}{b}\) is:
   \[
   \frac{a}{b} = \frac{3}{4}
   \]

The final answer is \(\boxed{\frac{3}{4}}\).