Find limit of the ratio of the area of the triangle formed by the origin and intersection points of the parabola `y=4x^2` and the line `y=a^2` to the area between the parabola and the line as `a` approaches to zero.

Answer

To solve the problem, we need to find the limit of the ratio of the area of a triangle formed by the origin and the intersection points of the parabola \( y = 4x^2 \) and the line \( y = a^2 \) to the area between the parabola and the line as \( a \) approaches zero.

### Step 1: Find the intersection points of the parabola and the line
The equations of the curves are:
- Parabola: \( y = 4x^2 \)
- Line: \( y = a^2 \)

To find the intersection points, we set the two equations equal to each other:
\[ 4x^2 = a^2 \]
This simplifies to:
\[ x^2 = \frac{a^2}{4} \]
Taking the square root gives us:
\[ x = \pm \frac{a}{2} \]
Thus, the intersection points are:
- \( A\left(-\frac{a}{2}, a^2\right) \)
- \( B\left(\frac{a}{2}, a^2\right) \)

### Step 2: Calculate the area of triangle \( OAB \)
The area \( A_1 \) of triangle \( OAB \) can be calculated using the formula for the area of a triangle given by vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points \( O(0, 0) \), \( A\left(-\frac{a}{2}, a^2\right) \), and \( B\left(\frac{a}{2}, a^2\right) \):
\[ A_1 = \frac{1}{2} \left| 0(a^2 - a^2) + \left(-\frac{a}{2}\right)(0 - a^2) + \left(\frac{a}{2}\right)(a^2 - 0) \right| \]
This simplifies to:
\[ A_1 = \frac{1}{2} \left| -\frac{a}{2}(-a^2) + \frac{a}{2}(a^2) \right| = \frac{1}{2} \left| \frac{a^3}{2} + \frac{a^3}{2} \right| = \frac{1}{2} \left| a^3 \right| = \frac{a^3}{4} \]

### Step 3: Calculate the area between the parabola and the line
The area \( A_2 \) between the parabola and the line from \( x = -\frac{a}{2} \) to \( x = \frac{a}{2} \) is given by:
\[ A_2 = \int_{-\frac{a}{2}}^{\frac{a}{2}} (a^2 - 4x^2) \, dx \]
Calculating the integral:
\[ A_2 = \int_{-\frac{a}{2}}^{\frac{a}{2}} a^2 \, dx - \int_{-\frac{a}{2}}^{\frac{a}{2}} 4x^2 \, dx \]
Calculating the first integral:
\[ \int_{-\frac{a}{2}}^{\frac{a}{2}} a^2 \, dx = a^2 \left[ x \right]_{-\frac{a}{2}}^{\frac{a}{2}} = a^2 \left( \frac{a}{2} - \left(-\frac{a}{2}\right) \right) = a^2 \cdot a = a^3 \]
Calculating the second integral:
\[ \int_{-\frac{a}{2}}^{\frac{a}{2}} 4x^2 \, dx = 4 \left[ \frac{x^3}{3} \right]_{-\frac{a}{2}}^{\frac{a}{2}} = 4 \left( \frac{(\frac{a}{2})^3}{3} - \frac{(-\frac{a}{2})^3}{3} \right) = 4 \cdot \frac{2a^3}{24} = \frac{a^3}{3} \]
Thus, we have:
\[ A_2 = a^3 - \frac{a^3}{3} = \frac{3a^3}{3} - \frac{a^3}{3} = \frac{2a^3}{3} \]

### Step 4: Find the limit of the ratio \( \frac{A_1}{A_2} \)
Now we can find the ratio:
\[ \frac{A_1}{A_2} = \frac{\frac{a^3}{4}}{\frac{2a^3}{3}} = \frac{a^3}{4} \cdot \frac{3}{2a^3} = \frac{3}{8} \]
Taking the limit as \( a \) approaches 0:
\[ \lim_{a \to 0} \frac{A_1}{A_2} = \frac{3}{8} \]

### Conclusion
The limit of the ratio of the area of the triangle to the area between the parabola and the line as \( a \) approaches zero is:
\[ \boxed{\frac{3}{8}} \]