The shortest distance between the line `y=x` and the curve `y^(2)=x-2` is:

Options

A. 2
B. `(7)/(8)`
C. `(7)/(4sqrt(2))`
D. `(11)/(4sqrt(2))`

Answer

To find the shortest distance between the line \( y = x \) and the curve \( y^2 = x - 2 \), we will follow these steps:

### Step 1: Parametrize the curve
The equation of the curve is \( y^2 = x - 2 \). We can express \( x \) in terms of \( y \):
\[ x = y^2 + 2 \]

### Step 2: Set up the distance formula
The distance \( D \) between a point on the curve \( (y^2 + 2, y) \) and a point on the line \( (x, x) \) can be expressed as:
\[ D = \sqrt{(x - (y^2 + 2))^2 + (x - y)^2} \]
Since \( x = y \) on the line, we can substitute \( x \) with \( y \):
\[ D = \sqrt{(y - (y^2 + 2))^2 + (y - y)^2} \]
This simplifies to:
\[ D = \sqrt{(y - y^2 - 2)^2} \]
Thus, we have:
\[ D = |y - y^2 - 2| \]

### Step 3: Minimize the distance
To find the shortest distance, we need to minimize \( D = |y - y^2 - 2| \). This requires finding the critical points of the function \( f(y) = y - y^2 - 2 \).

### Step 4: Find the derivative
Taking the derivative of \( f(y) \):
\[ f'(y) = 1 - 2y \]
Setting the derivative to zero to find critical points:
\[ 1 - 2y = 0 \implies y = \frac{1}{2} \]

### Step 5: Evaluate the function at the critical point
Now, we evaluate \( f(y) \) at \( y = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 - 2 = \frac{1}{2} - \frac{1}{4} - 2 = \frac{1}{4} - 2 = -\frac{7}{4} \]
Thus, the distance is:
\[ D = |-\frac{7}{4}| = \frac{7}{4} \]

### Step 6: Calculate the distance
The shortest distance \( D \) is:
\[ D = \frac{7}{4} \]

### Step 7: Finalize the distance formula
To express the distance in terms of the line equation, we need to consider the distance from the point on the curve to the line. The line can be expressed in the form \( ax + by + c = 0 \), where \( a = 1, b = -1, c = 0 \).

### Step 8: Substitute into the distance formula
The distance from a point \( (x_0, y_0) \) to the line \( ax + by + c = 0 \) is given by:
\[ D = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]
Substituting \( (x_0, y_0) = (y^2 + 2, y) \):
\[ D = \frac{|1(y^2 + 2) - 1(y)|}{\sqrt{1^2 + (-1)^2}} = \frac{|y^2 + 2 - y|}{\sqrt{2}} = \frac{|y^2 - y + 2|}{\sqrt{2}} \]

### Step 9: Find the minimum distance
We already found that \( y = \frac{1}{2} \) gives us the minimum distance. Thus, substituting back:
\[ D = \frac{7}{4\sqrt{2}} = \frac{7\sqrt{2}}{8} \]

### Conclusion
The shortest distance between the line \( y = x \) and the curve \( y^2 = x - 2 \) is:
\[ \frac{7\sqrt{2}}{8} \]