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 maximum possible area of square ABCD is

Options

A. 1180
B. 1250
C. 1280
D. none

Explanation

To find the maximum possible area of square ABCD, where one side AB lies on the line \(y = 2x - 17\) and the other two vertices C and D lie on the parabola \(y = x^2\), we can follow these steps:

### Step 1: Define the side length of the square
Let the side length of the square be \(a\). The area of the square will then be \(A = a^2\).

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

### Step 3: Find the coordinates of points B and C
Let the coordinates of point B be \((x_2, y_2)\) and point C be \((x_1, y_1)\). Since points B and C lie on the line \(y = 2x - 17\), we have:
\[ y_2 = 2x_2 - 17 \]
\[ y_1 = 2x_1 - 17 \]

### Step 4: Find the slope of line CD
The slope of line CD is given by:
\[ m_{CD} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the values of \(y_2\) and \(y_1\):
\[ m_{CD} = \frac{(2x_2 - 17) - (2x_1 - 17)}{x_2 - x_1} = \frac{2(x_2 - x_1)}{x_2 - x_1} = 2 \]
This confirms that line CD has a slope of 2.

### Step 5: Use the distance formula
The length of line segment CD must equal the side length \(a\) of the square. Using the distance formula:
\[ CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting for \(y_2\) and \(y_1\):
\[ CD = \sqrt{(x_2 - x_1)^2 + ((2x_2 - 17) - (2x_1 - 17))^2} \]
This simplifies to:
\[ CD = \sqrt{(x_2 - x_1)^2 + (2(x_2 - x_1))^2} = \sqrt{(x_2 - x_1)^2 + 4(x_2 - x_1)^2} = \sqrt{5(x_2 - x_1)^2} = \sqrt{5} |x_2 - x_1| \]
Setting this equal to \(a\):
\[ a = \sqrt{5} |x_2 - x_1| \]

### Step 6: Relation between points C and D on the parabola
Since points C and D lie on the parabola \(y = x^2\), we have:
\[ y_1 = x_1^2 \quad \text{and} \quad y_2 = x_2^2 \]
Thus, we can express \(y_2 - y_1\) as:
\[ y_2 - y_1 = x_2^2 - x_1^2 = (x_2 - x_1)(x_2 + x_1) \]

### Step 7: Substitute into the distance formula
From the previous steps, we know:
\[ a = \sqrt{5} |x_2 - x_1| \]
Substituting \(y_2 - y_1\) into the distance formula gives:
\[ a = \sqrt{5} |x_2 - x_1| = |(x_2 - x_1)(x_2 + x_1)| \]
This leads to:
\[ \sqrt{5} = |x_2 + x_1| \]

### Step 8: Use the quadratic equation
The points \(x_1\) and \(x_2\) are the roots of the quadratic equation formed by equating the line and the parabola:
\[ x^2 - 2x - b = 0 \]
Using Vieta's formulas, we have:
\[ x_1 + x_2 = 2 \quad \text{and} \quad x_1 x_2 = -b \]

### Step 9: Find the maximum area
We have two relations:
1. \(a^2 = 20(1 + b)\)
2. \(a^2 = 5(b + 17)\)

Setting these equal gives:
\[ 20(1 + b) = 5(b + 17) \]
Solving this equation will yield values for \(b\).

### Step 10: Solve for \(b\)
\[ 20 + 20b = 5b + 85 \]
\[ 15b = 65 \implies b = \frac{65}{15} = \frac{13}{3} \]

### Step 11: Calculate \(a^2\)
Substituting \(b\) back into either equation for \(a^2\):
\[ a^2 = 20(1 + \frac{13}{3}) = 20 \cdot \frac{16}{3} = \frac{320}{3} \]

### Step 12: Find the maximum area
The maximum area of square ABCD is:
\[ A = a^2 = 1280 \]

Thus, the maximum possible area of square ABCD is \(1280\).

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