How do you find the point on the parabola y² = 2x that is closest to the point (1, 4)?

To find the point on the parabola y² = 2x that is closest to the point (1, 4), we can use the method of minimizing the distance between a point on the parabola and the given point.

First, we express the distance D between a point (x, y) on the parabola and the point (1, 4) using the distance formula:

D = √((x - 1)² + (y - 4)²)

Since we want to minimize the distance, it’s easier to minimize the square of the distance (D²):

D² = (x - 1)² + (y - 4)²

Now, substitute y in terms of x from the parabola’s equation. From y² = 2x, we have:

y = ±√(2x)

For simplification, we can work with the positive branch of the parabola first. Thus, substituting y gives us:

D² = (x - 1)² + (√(2x) - 4)²

We need to minimize this function with respect to x. Expanding the expression results in:

D² = (x - 1)² + (2x - 8√(2x) + 16)

Next, we can combine the terms:

D² = x² - 2x + 1 + 2x - 8√(2x) + 16

Thus, we have:

D² = x² - 8√(2x) + 17

To find the minimum, we will take the derivative of with respect to x and set it to zero:

d(D²)/dx = 2x - (8 * (1/√(2x)) * (1/2√(2))) = 2x - 4√(2/x)

Setting the derivative equal to zero:

2x - 4√(2/x) = 0

Solving for x, we rearrange:

2x = 4√(2/x)
√(2/x) = x/2

Square both sides to eliminate the square root:

2/x = x²/4

Multiplying through by 4x gives:

8 = x³

This leads to:

x = 2

Now, substitute this value back into y² = 2x to find y:

y² = 2(2) = 4

Thus, y = ±2. We have two possible points on the parabola: (2, 2) and (2, -2).

Next, we compare the distances of these two points to (1, 4):

For point (2, 2):

D = √((2 - 1)² + (2 - 4)²) = √(1 + 4) = √5

For point (2, -2):

D = √((2 - 1)² + (-2 - 4)²) = √(1 + 36) = √37

Since √5 (approximately 2.24) is less than √37 (approximately 6.08), the point on the parabola y² = 2x that is closest to the point (1, 4) is:

(2, 2)

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