The triangle formed by the tangent to the parabola `y=x^(2)` at the point whose abscissa is k where `kin[1, 2]` the y-axis and the straight line `y=k^(2)` has greatest area if k is equal to

Options

A. 1
B. 3
C. 2
D. none of these

Answer

To solve the problem, we need to find the value of \( k \) such that the area of the triangle formed by the tangent to the parabola \( y = x^2 \) at the point where the abscissa is \( k \), the y-axis, and the line \( y = k^2 \) is maximized.

### Step-by-step Solution:

1. **Identify the point on the parabola:**
The point on the parabola \( y = x^2 \) at \( x = k \) is \( (k, k^2) \).

2. **Find the equation of the tangent line:**
The slope of the tangent to the parabola at the point \( (k, k^2) \) is given by the derivative \( \frac{dy}{dx} = 2x \). At \( x = k \), the slope is \( 2k \).
The equation of the tangent line at this point can be expressed using the point-slope form:
\[ y - k^2 = 2k(x - k) \]
Rearranging gives:
\[ y = 2kx - 2k^2 + k^2 = 2kx - k^2 \]
Thus, the equation of the tangent line is:
\[ y = 2kx - k^2 \]

3. **Find the intersection with the y-axis:**
To find the intersection with the y-axis, set \( x = 0 \):
\[ y = 2k(0) - k^2 = -k^2 \]
So the intersection point with the y-axis is \( (0, -k^2) \).

4. **Find the intersection with the line \( y = k^2 \):**
Set \( y = k^2 \) in the tangent equation:
\[ k^2 = 2kx - k^2 \]
Rearranging gives:
\[ 2kx = 2k^2 \implies x = k \quad (\text{if } k \neq 0) \]

5. **Determine the vertices of the triangle:**
The vertices of the triangle are:
- Point A: \( (0, -k^2) \) (intersection with y-axis)
- Point B: \( (k, k^2) \) (point on the parabola)
- Point C: \( (k, k^2) \) (intersection with the line \( y = k^2 \))

6. **Calculate the area of the triangle:**
The area \( A \) of the triangle formed by these points can be calculated using the formula for the area of a triangle:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the distance along the x-axis from the y-axis to the point \( (k, k^2) \), which is \( k \), and the height is the vertical distance from the point \( (k, k^2) \) to the point \( (0, -k^2) \), which is \( k^2 + k^2 = 2k^2 \).
Therefore, the area is:
\[ A = \frac{1}{2} \times k \times 2k^2 = k^3 \]

7. **Maximize the area:**
We need to maximize \( A = k^3 \) for \( k \in [1, 2] \). The function \( k^3 \) is increasing in this interval. Therefore, the maximum area occurs at the upper limit of the interval:
\[ k = 2 \]

### Conclusion:
The value of \( k \) for which the area of the triangle is maximized is \( k = 2 \).