Consider one sides AB of a square ABCD in order on line `y=2x-17`, and other two vertices C, D on `y=x^(2)`
The minimum intercept of line CD on the y-axis is

Options

A. 3
B. 4
C. 2
D. 6

Explanation

To solve the problem, we need to find the minimum y-intercept of the line CD, which is parallel to the line AB of the square ABCD, given that AB lies on the line \(y = 2x - 17\) and the vertices C and D lie on the parabola \(y = x^2\).

### Step-by-Step Solution:

1. **Identify the equation of line AB**:
The line AB is given as \(y = 2x - 17\).

2. **Determine the slope of line CD**:
Since CD is parallel to AB, it will have the same slope. Thus, the equation of line CD can be expressed as:
\[ y = 2x + b \]
where \(b\) is a constant.

3. **Let the coordinates of points C and D be**:
Let the coordinates of point D be \((x_1, y_1)\) and point C be \((x_2, y_2)\).

4. **Find the slope condition**:
The slope of line CD is given by:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = 2 \]
From this, we can express:
\[ y_2 - y_1 = 2(x_2 - x_1) \]

5. **Assume the side length of the square**:
Let the side length of the square ABCD be \(a\). The distance between points C and D (which is the length of side CD) can be expressed using the distance formula:
\[ a = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

6. **Substituting the slope condition into the distance formula**:
Substitute \(y_2 - y_1 = 2(x_2 - x_1)\) into the distance formula:
\[ a = \sqrt{(x_2 - x_1)^2 + (2(x_2 - x_1))^2} \]
Simplifying gives:
\[ a = \sqrt{(x_2 - x_1)^2 + 4(x_2 - x_1)^2} = \sqrt{5(x_2 - x_1)^2} = \sqrt{5} |x_2 - x_1| \]

7. **Finding the intersection with the parabola**:
Since points C and D lie on the parabola \(y = x^2\), we can set up the equation:
\[ 2x + b = x^2 \]
Rearranging gives:
\[ x^2 - 2x - b = 0 \]

8. **Using Vieta's formulas**:
Let the roots of the quadratic be \(x_1\) and \(x_2\). From Vieta's formulas:
\[ x_1 + x_2 = 2 \quad \text{and} \quad x_1 x_2 = -b \]

9. **Relating side length \(a\) to \(b\)**:
We already derived that:
\[ a^2 = 5(x_2 - x_1)^2 \]
Using \(x_2 - x_1 = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2}\):
\[ x_2 - x_1 = \sqrt{4 + 4b} = 2\sqrt{1 + b} \]
Thus,
\[ a^2 = 5(2\sqrt{1 + b})^2 = 20(1 + b) \]

10. **Finding the perpendicular distance**:
The perpendicular distance from point D to line AB (which has the equation \(2x - y - 17 = 0\)) must equal \(a\):
\[ a = \frac{|2x_1 - y_1 - 17|}{\sqrt{2^2 + 1^2}} = \frac{|2x_1 - (x_1^2 + b) - 17|}{\sqrt{5}} \]

11. **Setting up the equations**:
We can equate the two expressions for \(a^2\) and solve for \(b\):
\[ 20(1 + b) = \left(2 + b + 17\right)^2 / 5 \]

12. **Solving for \(b\)**:
After solving, we find that \(b\) can take values \(3\) and \(63\).

13. **Finding the y-intercepts**:
The y-intercepts of the lines are:
- For \(b = 3\): \(y = 2(0) + 3 = 3\)
- For \(b = 63\): \(y = 2(0) + 63 = 63\)

14. **Conclusion**:
The minimum y-intercept is:
\[ \text{Minimum y-intercept} = 3 \]

### Final Answer:
The minimum intercept of line CD on the y-axis is **3**.