Step 1: Understanding the Concept:
The shortest distance between a line and a curve occurs along the common normal.
Alternatively, we can express the distance from a generic point on the curve to the line as a function of a single parameter and then minimize this function.
Step 2: Key Formula or Approach:
Let a point on the curve $x = y^2$ be parameterized as $P(t^2, t)$.
The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
We will express this distance $d$ as a function of $t$ and minimize it using calculus.
Step 3: Detailed Explanation:
The equation of the line is $x - y + 1 = 0$.
Let $P(t^2, t)$ be an arbitrary point on the parabola $x = y^2$.
The distance $D$ from point $P$ to the line is:
\[ D(t) = \frac{|(1)(t^2) + (-1)(t) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|t^2 - t + 1|}{\sqrt{2}} \]
To minimize the distance, we need to minimize the function inside the absolute value, let's call it $f(t) = t^2 - t + 1$.
Since $t^2 - t + 1 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}$, it is always positive for all real $t$. Thus we can drop the absolute value for optimization.
Differentiating $f(t)$ with respect to $t$:
\[ f'(t) = 2t - 1 \]
Setting $f'(t) = 0$ for critical points:
\[ 2t - 1 = 0 \Rightarrow t = \frac{1}{2} \]
To confirm it's a minimum, find the second derivative: $f''(t) = 2>0$, so it is indeed a minimum.
Substitute $t = \frac{1}{2}$ back into the distance formula:
\[ D_{min} = \frac{\left| \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 \right|}{\sqrt{2}} \]
\[ D_{min} = \frac{\left| \frac{1}{4} - \frac{1}{2} + 1 \right|}{\sqrt{2}} = \frac{\left| \frac{1 - 2 + 4}{4} \right|}{\sqrt{2}} = \frac{\frac{3}{4}}{\sqrt{2}} \]
Rationalizing the denominator:
\[ D_{min} = \frac{3}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4 \times 2} = \frac{3\sqrt{2}}{8} \]
Step 4: Final Answer:
The shortest distance is $\frac{3\sqrt{2}}{8}$.