Key Concepts and Formulas

• Distance from a Point to a Line: The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
• Finding the Minimum Distance: The shortest distance between a curve and a line occurs where the tangent to the curve is parallel to the given line. This is because the distance function will have a minimum at this point.

Step-by-Step Solution

Step 1: Find the slope of the given line

The equation of the line is $y - x = 1$, which can be rewritten as $y = x + 1$. The slope of this line is $m = 1$. We need this slope to find the point on the curve where the tangent has the same slope.

Step 2: Find the derivative of the curve

The equation of the curve is $x = y^2$. Differentiating both sides with respect to $x$, we get: $1 = 2y \frac{dy}{dx}$ Therefore, the derivative $\frac{dy}{dx}$ is: $\frac{dy}{dx} = \frac{1}{2y}$ This derivative represents the slope of the tangent to the curve at any point $(x, y)$.

Step 3: Find the point on the curve where the tangent is parallel to the line

For the tangent to be parallel to the line, their slopes must be equal. Therefore, we set the derivative equal to the slope of the line: $\frac{1}{2y} = 1$ Solving for $y$, we get: $y = \frac{1}{2}$ Now, substitute this value of $y$ into the equation of the curve $x = y^2$ to find the corresponding $x$-coordinate: $x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ So, the point on the curve closest to the line is $\left(\frac{1}{4}, \frac{1}{2}\right)$.

Step 4: Calculate the distance from the point to the line

The equation of the line is $y - x = 1$, which can be rewritten as $x - y + 1 = 0$. Now, we use the distance formula to find the distance from the point $\left(\frac{1}{4}, \frac{1}{2}\right)$ to the line $x - y + 1 = 0$: $d = \frac{\left|1\left(\frac{1}{4}\right) - 1\left(\frac{1}{2}\right) + 1\right|}{\sqrt{1^2 + (-1)^2}}$ $d = \frac{\left|\frac{1}{4} - \frac{1}{2} + 1\right|}{\sqrt{2}}$ $d = \frac{\left|\frac{1 - 2 + 4}{4}\right|}{\sqrt{2}}$ $d = \frac{\left|\frac{3}{4}\right|}{\sqrt{2}}$ $d = \frac{3}{4\sqrt{2}}$ $d = \frac{3\sqrt{2}}{4\sqrt{2}\sqrt{2}}$ $d = \frac{3\sqrt{2}}{8}$

Common Mistakes & Tips

• Incorrect Differentiation: Make sure to differentiate the curve equation correctly, especially when dealing with implicit differentiation.
• Forgetting Absolute Value: Remember to use the absolute value in the distance formula to ensure the distance is positive.
• Rationalizing the Denominator: It's generally good practice to rationalize the denominator to match the answer format.

Summary

We found the shortest distance between the line $y - x = 1$ and the curve $x = y^2$ by first finding the point on the curve where the tangent is parallel to the line. This involved finding the derivative of the curve, setting it equal to the slope of the line, and solving for the coordinates of the point. Then, we used the point-to-line distance formula to calculate the shortest distance, which is $\frac{3\sqrt{2}}{8}$.

Final Answer The final answer is $\boxed{\frac{3\sqrt{2}}{8}}$, which corresponds to option (D).