Find the shortest distance between the line `x - y +1 = 0` and the curve `y^2 = x.`

Options

A. `(3sqrt(2))/(8)`
B. `(2sqrt(3))/(8)`
C. `(3sqrt(2))/(5)`
D. `(sqrt(3))/(4)`

Answer

To find the shortest distance between the line \( x - y + 1 = 0 \) and the curve \( y^2 = x \), we can follow these steps:

### Step 1: Parameterize the Curve
The curve given is \( y^2 = x \). We can express \( x \) and \( y \) in terms of a parameter \( t \):
- Let \( y = t \)
- Then, \( x = t^2 \)

### Step 2: Write the Distance Formula
The distance \( D \) from a point \( (x_0, y_0) \) to a line \( ax + by + c = 0 \) is given by the formula:
\[ D = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]
For our line \( x - y + 1 = 0 \), we have:
- \( a = 1 \)
- \( b = -1 \)
- \( c = 1 \)

### Step 3: Substitute the Parameterization into the Distance Formula
Substituting \( x_0 = t^2 \) and \( y_0 = t \) into the distance formula:
\[ D = \frac{|1(t^2) - 1(t) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|t^2 - t + 1|}{\sqrt{2}} \]

### Step 4: Simplify the Expression
We can simplify the expression for distance:
\[ D = \frac{|t^2 - t + 1|}{\sqrt{2}} \]

### Step 5: Find the Minimum Distance
To find the shortest distance, we need to minimize \( |t^2 - t + 1| \). We can find the critical points by differentiating the expression inside the absolute value:
\[ f(t) = t^2 - t + 1 \]
Calculating the derivative:
\[ f'(t) = 2t - 1 \]
Setting the derivative to zero to find critical points:
\[ 2t - 1 = 0 \implies t = \frac{1}{2} \]

### Step 6: Evaluate the Function at the Critical Point
Now we evaluate \( f(t) \) at \( t = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 7: Calculate the Minimum Distance
Now substituting back into the distance formula:
\[ D_{\text{min}} = \frac{|f\left(\frac{1}{2}\right)|}{\sqrt{2}} = \frac{\frac{3}{4}}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8} \]

### Conclusion
The shortest distance between the line \( x - y + 1 = 0 \) and the curve \( y^2 = x \) is \( \frac{3\sqrt{2}}{8} \).

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