To solve the problem, we need to find the value of \( k \) for which the area of the triangle formed by the tangent to the parabola \( y = x^2 \) at the point with abscissa \( k \), the y-axis, and the line \( y = k^2 \) is maximized, given that \( k \) is in the interval \([1, 2]\).
### Step-by-Step Solution:
1. **Find the point of tangency on the parabola:**
The point on the parabola where the abscissa is \( k \) is \( (k, k^2) \).
2. **Find the equation of the tangent line:**
The slope of the tangent to the parabola \( y = x^2 \) at the point \( (k, k^2) \) is given by the derivative \( \frac{dy}{dx} = 2k \). Using the point-slope form of the line, the equation of the tangent line is:
\[
y - k^2 = 2k(x - k)
\]
Simplifying this, we get:
\[
y = 2kx - 2k^2 + k^2 = 2kx - k^2
\]
Therefore, the equation of the tangent line is:
\[
y = 2kx - k^2
\]
3. **Find the points of intersection with the y-axis and the line \( y = k^2 \):**
- **Intersection with the y-axis (where \( x = 0 \)):**
\[
y = 2k(0) - k^2 = -k^2
\]
So, the point of intersection with the y-axis is \( A(0, -k^2) \).
- **Intersection with the line \( y = k^2 \):**
Set \( y = k^2 \) in the tangent equation:
\[
k^2 = 2kx - k^2
\]
Rearranging gives:
\[
2kx = 2k^2 \implies x = k
\]
Thus, the point of intersection is \( B(k, k^2) \).
4. **Find the third vertex of the triangle:**
The third vertex is the intersection of the line \( y = k^2 \) with the y-axis, which is \( C(0, k^2) \).
5. **Determine the vertices of the triangle:**
The vertices of the triangle are:
- \( A(0, -k^2) \)
- \( B(k, k^2) \)
- \( C(0, k^2) \)
6. **Calculate the area of the triangle using the formula:**
The area \( A \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is given by:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the points:
\[
A = \frac{1}{2} \left| 0(k^2 - k^2) + k(k^2 - (-k^2)) + 0(-k^2 - k^2) \right|
\]
Simplifying this:
\[
A = \frac{1}{2} \left| k(2k^2) \right| = \frac{1}{2} \cdot 2k^3 = k^3
\]
7. **Maximize the area \( A = k^3 \) over the interval \( [1, 2] \):**
The function \( k^3 \) is increasing in the interval \( [1, 2] \). Therefore, the maximum area occurs at the endpoint \( k = 2 \).
### Conclusion:
The value of \( k \) for which the area of the triangle is maximized is:
\[
\boxed{2}
\]
