Key Concept: Shortest Distance Between a Line and a Convex Curve

The shortest distance between a straight line and a non-intersecting convex curve (like a parabola, ellipse, or hyperbola that does not cross the line) is found using a specific geometric principle. This principle states that the shortest distance occurs at a point on the curve where the tangent to the curve is parallel to the given straight line.

Why this principle works: Imagine a series of parallel lines to the given line, moving closer and closer to the curve. The first such line that touches the curve will be tangent to it at the point of closest approach. Since this tangent line is parallel to the given line, the line segment connecting the point of tangency on the curve to the given line, representing the shortest distance, will be perpendicular to both the given line and this tangent. This means the normal to the curve at the point of closest approach is perpendicular to the given line.

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Step-by-Step Solution

1. Identify and Analyze the Given Line and Curve

We are provided with two equations:

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

Let's analyze these:

• The line $y=x$ is a straight line passing through the origin $(0,0)$ with a slope of $1$.
• The curve $y^2 = x - 2$ can be rewritten as $x = y^2 + 2$. This is a parabola opening to the right, with its vertex at $(2, 0)$.

Visual Check: Since the parabola opens to the right from $(2,0)$ and the line $y=x$ passes through the origin, it's clear that the line and the parabola do not intersect. This confirms that a shortest distance exists and our key concept is applicable.

2. Determine the Slope of the Given Line

The equation of the line is $y = x$. This is in the slope-intercept form $y = mx + c$, where $m$ is the slope. Comparing $y=x$ with $y=mx+c$: The slope of the given line $L$ is $m_L = 1$.

Why we need this: According to our key concept, the tangent to the parabola at the point of closest approach must have this same slope.

3. Find the Slope of the Tangent to the Parabola

To find the slope of the tangent to the parabola $y^2 = x - 2$ at any point $(x, y)$ on the curve, we need to calculate its derivative $\frac{dy}{dx}$. We will use implicit differentiation with respect to $x$: $\frac{d}{dx}(y^2) = \frac{d}{dx}(x - 2)$ Applying the chain rule for $y^2$ and differentiating the right side: $2y \frac{dy}{dx} = 1 - 0$ $2y \frac{dy}{dx} = 1$ Now, we solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{2y}$ This expression gives the slope of the tangent to the parabola at any point $(x, y)$ on the curve.

Why implicit differentiation: The equation of the parabola is not explicitly solved for $y$ in terms of $x$. Implicit differentiation allows us to find $\frac{dy}{dx}$ by treating $y$ as a function of $x$.

4. Determine the Coordinates of the Point of Closest Approach $P(x_P, y_P)$

For the tangent at the point $P(x_P, y_P)$ on the parabola to be parallel to the line $y=x$, their slopes must be equal. So, we set the slope of the tangent at $P$ equal to the slope of the line $L$: $\left(\frac{dy}{dx}\right)_P = m_L$ $\frac{1}{2y_P} = 1$ Solving for $y_P$: $2y_P = 1 \implies y_P = \frac{1}{2}$ Now that we have the y-coordinate of the point $P$, we substitute this $y_P$ value back into the parabola's equation $y^2 = x - 2$ to find the corresponding $x_P$: $(y_P)^2 = x_P - 2$ $\left(\frac{1}{2}\right)^2 = x_P - 2$ $\frac{1}{4} = x_P - 2$ To find $x_P$, we add 2 to both sides: $x_P = 2 + \frac{1}{4}$ $x_P = \frac{8}{4} + \frac{1}{4} = \frac{9}{4}$ Thus, the point $P$ on the parabola where the tangent is parallel to $y=x$ is $P\left(\frac{9}{4}, \frac{1}{2}\right)$. This is the point on the parabola closest to the line $y=x$.

Why we equate slopes: This is the direct application of our key concept. Finding $P$ is crucial because the shortest distance will be the perpendicular distance from this specific point $P$ to the given line.

5. Calculate the Shortest Distance

The shortest distance is the perpendicular distance from the point $P\left(\frac{9}{4}, \frac{1}{2}\right)$ to the line $y=x$.

First, we rewrite the line equation $y=x$ into the general form $Ax + By + C = 0$: $x - y + 0 = 0$ From this, we identify the coefficients: $A=1$, $B=-1$, and $C=0$. The formula for the perpendicular distance $D$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is: $D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ Now, substitute the coordinates of point $P(x_1, y_1) = \left(\frac{9}{4}, \frac{1}{2}\right)$ and the coefficients of the line into the formula: $D = \frac{\left|1 \cdot \left(\frac{9}{4}\right) + (-1) \cdot \left(\frac{1}{2}\right) + 0\right|}{\sqrt{(1)^2 + (-1)^2}}$ $D = \frac{\left|\frac{9}{4} - \frac{1}{2}\right|}{\sqrt{1+1}}$ Simplify the expression inside the absolute value in the numerator: $\frac{9}{4} - \frac{1}{2} = \frac{9}{4} - \frac{2}{4} = \frac{7}{4}$ Substitute this back into the distance formula: $D = \frac{\left|\frac{7}{4}\right|}{\sqrt{2}}$ Since $\frac{7}{4}$ is positive, the absolute value is simply $\frac{7}{4}$: $D = \frac{\frac{7}{4}}{\sqrt{2}}$ $D = \frac{7}{4\sqrt{2}}$

Why the perpendicular distance formula: Once the closest point $P$ is found, the shortest distance from $P$ to the line is by definition the perpendicular distance. This formula is standard for calculating it.

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Important Tips and Common Pitfalls

• Understanding the Concept: Always start by clearly understanding why the tangent must be parallel. A visual sketch can be very helpful.
• Implicit Differentiation: Be careful with the chain rule when differentiating terms involving $y$. Remember that $\frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx}$.
• Algebraic Accuracy: Double-check all arithmetic, especially when dealing with fractions while solving for $x_P, y_P$ and in the final distance calculation. A small error here can lead to an incorrect option.
• Distance Formula: Ensure you use the correct formula for the perpendicular distance from a point to a line. Convert the line equation to $Ax+By+C=0$ form correctly before plugging in values. Pay attention to the absolute value in the numerator.
• Rationalization: Sometimes, options might have the denominator rationalized (e.g., $\frac{7\sqrt{2}}{8}$). While $\frac{7}{4\sqrt{2}}$ is a valid answer, be prepared to rationalize if your calculated answer isn't directly in the options. In this case, $\frac{7}{4\sqrt{2}} = \frac{7\sqrt{2}}{4 \cdot 2} = \frac{7\sqrt{2}}{8}$. Both forms are equivalent.

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Summary of Solution

To find the shortest distance between the line $y=x$ and the parabola $y^2=x-2$, we utilized the principle that the tangent to the parabola at the point of closest approach must be parallel to the given line.

1. We identified the slope of the line $y=x$ as $m_L = 1$.
2. We found the general slope of the tangent to the parabola by implicit differentiation: $\frac{dy}{dx} = \frac{1}{2y}$.
3. By equating these slopes, $\frac{1}{2y_P} = 1$, we found the y-coordinate of the closest point $y_P = \frac{1}{2}$. Substituting this back into the parabola's equation, we found $x_P = \frac{9}{4}$. So, the point of closest approach is $P\left(\frac{9}{4}, \frac{1}{2}\right)$.
4. Finally, we calculated the perpendicular distance from this point $P$ to the line $x-y=0$ using the distance formula, which resulted in $D = \frac{7}{4\sqrt{2}}$.

The final answer is $\boxed{\text{A}}$.