1. Understanding the Concept: Shortest Distance Between a Curve and a Line

The problem asks for the shortest distance between a given line and a given curve. This is a fundamental concept in coordinate geometry and differential calculus.

The key principle for finding the shortest distance between a continuous curve and a straight line is as follows: The shortest distance occurs at a point on the curve where the tangent to the curve is parallel to the given line. Imagine taking a line parallel to the given line and moving it towards the curve. The very first point where this moving line just touches the curve (i.e., becomes tangent to the curve) is the point on the curve closest to the original line.

Once we identify this specific point on the curve, say $P(x_0, y_0)$, the shortest distance is simply the perpendicular distance from this point $P$ to the given line.

The formula for the perpendicular distance from a point $P(x_0, y_0)$ to a line $Ax + By + C = 0$ is: $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

A Crucial Initial Check: Intersection! Before attempting to apply derivatives, it is absolutely essential to first check if the line and the curve intersect. If they intersect at one or more points, the shortest distance between them is 0. This check can save significant time and prevent unnecessary calculations, especially if "0" is an option.

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2. Analyzing the Given Problem

We are given:

• The line $L: x - y = 1$
• The curve $C: x^2 = 2y$

Let's prepare these equations for analysis:

• Line $L$: We can rewrite $x - y = 1$ in the slope-intercept form $y = mx + c$ to easily identify its slope. $y = x - 1$ The slope of the line $L$ is $m_L = 1$.
• Curve $C$: The equation $x^2 = 2y$ (or $y = \frac{1}{2}x^2$) represents a parabola opening upwards, with its vertex at the origin $(0,0)$.

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3. Step-by-Step Solution

Step 3.1: Check for Intersection Between the Line and the Curve (The Most Important First Step)

To determine if the line and the parabola intersect, we look for common points $(x, y)$ that satisfy both equations. We do this by substituting the expression for $y$ from the line equation into the parabola's equation.

Substitute $y = x - 1$ (from the line $L$) into $x^2 = 2y$ (the parabola $C$): $x^2 = 2(x - 1)$ $x^2 = 2x - 2$ Rearrange this into a standard quadratic equation: $x^2 - 2x + 2 = 0$

Now, we analyze the nature of the roots of this quadratic equation using its discriminant, $\Delta = b^2 - 4ac$. For $x^2 - 2x + 2 = 0$, we have $a = 1$, $b = -2$, and $c = 2$. $\Delta = (-2)^2 - 4(1)(2)$ $\Delta = 4 - 8$ $\Delta = -4$

Since the discriminant $\Delta < 0$, the quadratic equation has no real solutions for $x$. This means there are no points $(x, y)$ that satisfy both the line and the parabola equations simultaneously. Conclusion: The line and the parabola do not intersect. Therefore, the shortest distance between them is not 0. This directly tells us that option (A) is incorrect.

Step 3.2: Find the Slope of the Tangent to the Curve

To apply the shortest distance condition (tangent parallel to the line), we first need a way to calculate the slope of the tangent to the parabola at any general point $(x, y)$ on it. We achieve this using differentiation.

Given the curve $C: x^2 = 2y$. Differentiate both sides of the equation with respect to $x$: $\frac{d}{dx}(x^2) = \frac{d}{dx}(2y)$ $2x = 2 \frac{dy}{dx}$ Solve for $\frac{dy}{dx}$, which represents the slope of the tangent, $m_T$: $\frac{dy}{dx} = x$ So, the slope of the tangent to the parabola at any point $(x_0, y_0)$ on the curve is $m_T = x_0$.

Step 3.3: Apply the Shortest Distance Condition to Find the Closest Point

For the shortest distance, the tangent to the parabola at the closest point $P(x_0, y_0)$ must be parallel to the given line $L$. This means their slopes must be equal.

From Step 2, the slope of the line $L$ is $m_L = 1$. From Step 3.2, the slope of the tangent at $P(x_0, y_0)$ is $m_T = x_0$.

Setting the slopes equal: $m_T = m_L$ $x_0 = 1$

Now that we have the $x$-coordinate of the closest point, $x_0 = 1$, we need to find its corresponding $y$-coordinate. Since this point lies on the parabola, we substitute $x_0 = 1$ back into the parabola's equation ($x^2 = 2y$): $(1)^2 = 2y_0$ $1 = 2y_0$ $y_0 = \frac{1}{2}$

Thus, the point on the parabola closest to the line $x - y = 1$ is $P\left(1, \frac{1}{2}\right)$.

Step 3.4: Calculate the Shortest Distance

Finally, we calculate the perpendicular distance from the closest point $P\left(1, \frac{1}{2}\right)$ to the given line $x - y = 1$.

First, rewrite the line equation in the standard form $Ax + By + C = 0$: $x - y - 1 = 0$ From this, we identify $A = 1$, $B = -1$, and $C = -1$. The point is $(x_0, y_0) = \left(1, \frac{1}{2}\right)$.

Using the distance formula $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$: $D = \frac{|(1)(1) + (-1)\left(\frac{1}{2}\right) + (-1)|}{\sqrt{(1)^2 + (-1)^2}}$ $D = \frac{\left|1 - \frac{1}{2} - 1\right|}{\sqrt{1 + 1}}$ $D = \frac{\left|-\frac{1}{2}\right|}{\sqrt{2}}$ $D = \frac{\frac{1}{2}}{\sqrt{2}}$ $D = \frac{1}{2\sqrt{2}}$

To rationalize the denominator (a standard practice for simplifying expressions and matching options), multiply the numerator and denominator by $\sqrt{2}$: $D = \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$

Both $\frac{1}{2\sqrt{2}}$ and $\frac{\sqrt{2}}{4}$ are equivalent forms of the shortest distance. Comparing with the given options, $\frac{1}{2\sqrt{2}}$ matches option (B).

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4. Important Tips and Common Pitfalls

• Always Check for Intersection First: This is the most common pitfall. If the curve and line intersect, the shortest distance is 0. Failing to check this can lead to unnecessary calculations and an incorrect answer if 0 is an option.
• Understand the Condition: Remember that for the shortest distance between a curve and a line, the tangent to the curve at the closest point must be parallel to the line. This means their slopes are equal. Do not confuse this with conditions for perpendicularity.
• Derivative Accuracy: Be meticulous when differentiating to find the slope of the tangent. Any mistake here will propagate through the rest of the solution.
• Point-to-Line Distance Formula: Ensure you use the correct formula and substitute values accurately, paying close attention to signs and the absolute value.
• Simplification and Rationalization: Always simplify your final answer and rationalize the denominator if required, to match the format of the options.

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5. Summary and Key Takeaway

The shortest distance between the line $x - y = 1$ and the parabola $x^2 = 2y$ is $\frac{1}{2\sqrt{2}}$ (or $\frac{\sqrt{2}}{4}$). This was determined by following a systematic approach:

1. Crucially, we first verified that the line and curve do not intersect, confirming the shortest distance is not 0.
2. We found the point on the parabola where its tangent is parallel to the given line. This point was $P\left(1, \frac{1}{2}\right)$.
3. Finally, we calculated the perpendicular distance from this point $P$ to the given line.

The calculated shortest distance is $\frac{1}{2\sqrt{2}}$, which corresponds to option (B). The initial "Correct Answer: A" (0) was incorrect because the line and parabola do not intersect. This problem strongly emphasizes the importance of the initial intersection check.