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

Options

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

Explanation

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

### Step 1: Write the equations of the line and the curve
The line can be expressed as:
\[ x - y = 0 \]
The curve is given by:
\[ y^2 = x - 2 \]

### Step 2: Parameterize the curve
Let \( y = t \). Then, substituting into the curve's equation:
\[ y^2 = t^2 \implies x = t^2 + 2 \]
Thus, a point on the curve can be represented as \( P(t) = (t^2 + 2, t) \).

### Step 3: Use the distance formula
The distance \( d \) from the point \( P(t) \) to the line \( x - y = 0 \) can be calculated using the formula:
\[ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]
where \( (x_0, y_0) \) is the point on the curve, and \( ax + by + c = 0 \) is the line equation. Here, \( a = 1, b = -1, c = 0 \).

Substituting \( P(t) \) into the formula:
\[ d = \frac{|(1)(t^2 + 2) + (-1)(t) + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{|t^2 + 2 - t|}{\sqrt{2}} = \frac{|t^2 - t + 2|}{\sqrt{2}} \]

### Step 4: Minimize the distance
To find the shortest distance, we need to minimize \( |t^2 - t + 2| \). The expression inside the absolute value is a quadratic function:
\[ f(t) = t^2 - t + 2 \]
The vertex of a quadratic \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \):
\[ t = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]

### Step 5: Evaluate the function at the vertex
Now, substitute \( t = \frac{1}{2} \) into \( f(t) \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} = \frac{7}{4} \]

### Step 6: Calculate the minimum distance
Thus, the minimum distance \( d \) is:
\[ d = \frac{7/4}{\sqrt{2}} = \frac{7}{4\sqrt{2}} = \frac{7\sqrt{2}}{8} \]

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