1. Key Concept: Finding the Closest Point on a Curve to a Straight Line

When faced with the task of finding a point on a given curve that is closest to a specific straight line, a fundamental geometric principle, deeply rooted in calculus, provides an elegant solution.

The core idea is this: The point on the curve that minimizes the perpendicular distance to the line is characterized by the fact that the tangent to the curve at that point must be parallel to the given straight line.

• Why is this true? Imagine we have our given line, let's call it $L$. Now, consider drawing a family of lines, all parallel to $L$. As we move these parallel lines closer and closer to the curve, the very first line that just "touches" the curve (i.e., is tangent to it) will contain the point on the curve that is closest to the original line $L$. Since all lines in this family are parallel to $L$, the tangent line at this closest point must also be parallel to $L$. If the tangent were not parallel, it would imply we could find a point on the curve with a tangent closer to being parallel to $L$, which would lead to a shorter distance, contradicting our assumption of the closest point.
• Another perspective (using perpendicularity): The shortest distance between a point $P$ and a line $L$ is always measured along a segment perpendicular to $L$. If $P$ is the closest point on the curve to $L$, and $T$ is the tangent to the curve at $P$, then the line segment connecting $P$ to $L$ (representing the shortest distance) must be perpendicular to $L$. Since $T$ is parallel to $L$, the shortest distance segment must also be perpendicular to $T$. This geometric relationship confirms that the tangent at the closest point must be parallel to the given line.

Therefore, the strategy to find such a point involves the following steps:

1. Determine the slope of the given straight line.
2. Find a general expression for the slope of the tangent to the curve at any point $(x, y)$ using differentiation ($\frac{dy}{dx}$).
3. Equate these two slopes (the slope of the given line and the slope of the tangent at the closest point). This equality will allow us to solve for the x-coordinate of the closest point.
4. Substitute this x-coordinate back into the original curve's equation to find the corresponding y-coordinate.

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2. Problem Setup and Goal

We are given:

• The equation of a parabola: $y = x^2 + 4$
• The equation of a straight line: $y = 4x - 1$

Our objective is to find the coordinates of point $P(x_P, y_P)$ that lies on the parabola $y = x^2 + 4$ and is closest to the straight line $y = 4x - 1$.

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3. Step-by-Step Solution

Step 1: Determine the Slope of the Given Line

The equation of the straight line is given in the slope-intercept form, $y = mx + c$, where $m$ represents the slope of the line and $c$ is the y-intercept. Our given line is: $y = 4x - 1$ By comparing this to $y = mx + c$, we can directly identify the slope of the line: $m_{\text{line}} = 4$ This slope tells us the 'steepness' and direction of the line we're trying to find the closest point to.

Step 2: Find the General Slope of the Tangent to the Parabola

To find the slope of the tangent line at any arbitrary point $(x, y)$ on the parabola $y = x^2 + 4$, we utilize differential calculus. The derivative $\frac{dy}{dx}$ gives us the instantaneous rate of change of $y$ with respect to $x$, which is precisely the slope of the tangent line at that point. Given the parabola: $y = x^2 + 4$ Differentiating both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^2 + 4)$ Applying the power rule for differentiation ($\frac{d}{dx}(x^n) = nx^{n-1}$) and the rule for differentiating a constant ($\frac{d}{dx}(c) = 0$): $\frac{dy}{dx} = 2x + 0$ $\frac{dy}{dx} = 2x$ So, the general expression for the slope of the tangent to the parabola at any point $(x, y)$ is $m_{\text{tangent}} = 2x$. If $P(x_P, y_P)$ is the point we are looking for, then the slope of the tangent at $P$ is $2x_P$.

Step 3: Apply the Closest Point Condition and Solve for the x-coordinate of P ($x_P$)

As established in our key concept, for point $P$ to be the closest point on the parabola to the line, the tangent to the parabola at $P$ must be parallel to the given line. Parallel lines have identical slopes. Therefore, we equate the slope of the tangent at $P$ to the slope of the given line: $m_{\text{tangent}} = m_{\text{line}}$ Substituting the expressions we found in Step 1 and Step 2: $2x_P = 4$ Now, we solve this simple linear equation for $x_P$: $x_P = \frac{4}{2}$ $x_P = 2$ This value, $x_P = 2$, is the x-coordinate of the specific point on the parabola where the tangent is parallel to the line $y = 4x - 1$. This is the x-coordinate of our closest point $P$.

Step 4: Calculate the Corresponding y-coordinate of P ($y_P$)

Since the point $P(x_P, y_P)$ lies on the parabola $y = x^2 + 4$, its coordinates must satisfy the parabola's equation. We substitute the calculated value of $x_P = 2$ back into the parabola's equation to find the corresponding $y_P$: $y_P = (x_P)^2 + 4$ $y_P = (2)^2 + 4$ $y_P = 4 + 4$ $y_P = 8$ Thus, the coordinates of the point P are $(2, 8)$.

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4. Verification and Common Pitfalls

The coordinates of the point P are $(2, 8)$. Let's check the options provided. Option (B) is (2, 8).

• Common Mistake 1: Substituting into the Wrong Equation. A very frequent error is to substitute the calculated $x_P$ value (which is $2$ in this case) back into the line's equation ($y = 4x - 1$) instead of the parabola's equation. Remember, point P lies on the parabola, so its coordinates must satisfy the parabola's equation. If you mistakenly substitute $x=2$ into $y=4x-1$, you would get $y=4(2)-1 = 8-1=7$, leading to the incorrect point $(2,7)$. This point $(2,7)$ lies on the line, not on the parabola. Always verify that the final point lies on the original curve.
• Common Mistake 2: Forgetting the y-coordinate. Students sometimes find $x_P$ and stop, forgetting that the question asks for the coordinates of P, which means both x and y.
• Tip for Visualization: It's always helpful to visualize the problem. The parabola $y = x^2 + 4$ is an upward-opening parabola with its vertex at $(0,4)$. The line $y = 4x - 1$ has a positive slope and a negative y-intercept. If you were to sketch them, you would observe that the point $(2,8)$ is indeed on the parabola, and the tangent at this point would visually appear parallel to the given line.
• Generalizability: This method of equating slopes (using differentiation) is a powerful and general technique. It can be applied to find the closest point on any differentiable curve to a given straight line.
• Alternative (Less Efficient) Method: One could theoretically use the distance formula between a generic point $(x, x^2+4)$ on the parabola and the line $4x - y - 1 = 0$. The distance $D$ would be given by the formula $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. For our case, $D = \frac{|4x - (x^2+4) - 1|}{\sqrt{4^2 + (-1)^2}} = \frac{|-x^2 + 4x - 5|}{\sqrt{17}}$. One would then need to minimize this distance $D$ (or more simply, minimize $D^2$) by setting its derivative with respect to $x$ to zero. This method is mathematically sound but often involves more complex algebraic manipulations and differentiation than the slope-equating method, making it generally less efficient for this type of problem.

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5. Summary and Key Takeaway

By applying the fundamental geometric principle that the tangent to the parabola at the closest point must be parallel to the given line, we efficiently determined the coordinates of point P. We first calculated the slope of the line, then derived the general expression for the slope of the parabola's tangent using differentiation. Equating these slopes allowed us to solve for the x-coordinate of P. Finally, substituting this x-coordinate back into the parabola's equation yielded the corresponding y-coordinate.

The coordinates of the point P on the parabola $y = x^2 + 4$ that is closest to the line $y = 4x - 1$ are $(2, 8)$.

The final answer is $\boxed{\text{(2, 8)}}$.