Key Concepts and Formulas

• Shortest Distance: The shortest distance between a line and a curve occurs where the tangent to the curve is parallel to the line. The common normal at the point of tangency represents this shortest distance.
• Slope of a Line: The slope of a line in the form $y = mx + c$ is $m$.
• Derivative for Slope: The derivative of a curve $x = f(y)$ with respect to $y$, i.e., $\frac{dx}{dy}$, gives the reciprocal of the slope of the tangent at a point on the curve. That is, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.
• Distance Formula: The perpendicular distance between a point $(x_1, y_1)$ and a line $ax + by + c = 0$ is given by $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.

Step-by-Step Solution

Step 1: Identify the Equations of the Line and the Curve

We are given:

• The line: $y - x = 1$, which can be rewritten as $y = x + 1$.
• The curve: $x = y^2$.

Step 2: Determine the Slope of the Given Line

The equation of the line is $y = x + 1$. Comparing this with the slope-intercept form $y = mx + c$, we find that the slope of the line is $m = 1$.

Step 3: Find the Derivative of the Curve

We have $x = y^2$. Differentiating both sides with respect to $y$, we get: $\frac{dx}{dy} = 2y$ The slope of the tangent to the curve is given by $\frac{dy}{dx}$. Therefore, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2y}$

Step 4: 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, $\frac{1}{2y} = 1$ $2y = 1$ $y = \frac{1}{2}$

Step 5: Find the x-coordinate of the Point on the Curve

Substitute $y = \frac{1}{2}$ into the equation of the curve $x = y^2$: $x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ So, the point on the curve is $\left(\frac{1}{4}, \frac{1}{2}\right)$.

Step 6: Rewrite the Equation of the Line in the General Form

The given line is $y - x = 1$, which can be rewritten as $x - y + 1 = 0$.

Step 7: Calculate the Shortest Distance

The shortest distance between the point $\left(\frac{1}{4}, \frac{1}{2}\right)$ and the line $x - y + 1 = 0$ is given by the perpendicular distance formula: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = \frac{\left|1\left(\frac{1}{4}\right) - 1\left(\frac{1}{2}\right) + 1\right|}{\sqrt{1^2 + (-1)^2}}$ $d = \frac{\left|\frac{1}{4} - \frac{1}{2} + 1\right|}{\sqrt{2}} = \frac{\left|\frac{1 - 2 + 4}{4}\right|}{\sqrt{2}} = \frac{\left|\frac{3}{4}\right|}{\sqrt{2}} = \frac{3}{4\sqrt{2}}$ $d = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{4\sqrt{2}\sqrt{2}} = \frac{3\sqrt{2}}{4(2)} = \frac{3\sqrt{2}}{8}$

Common Mistakes & Tips

• Mistake: Forgetting to take the reciprocal of $\frac{dx}{dy}$ to get the slope $\frac{dy}{dx}$.
• Tip: Remember that the shortest distance occurs where the tangent to the curve is parallel to the given line.
• Tip: Ensure the line equation is in the form $ax + by + c = 0$ before applying the perpendicular distance formula.

Summary

To find the shortest distance between the line $y-x=1$ and the curve $x=y^2$, we first found the point on the curve where the tangent is parallel to the line. This was achieved by equating the slope of the line to the derivative of the curve. Then, we used the point-to-line distance formula to calculate the perpendicular distance between the point and the line, which represents the shortest distance.

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