To find the point on the parabola defined by the equation y² = 2x that is closest to the point (1, 4), we can follow a systematic approach using calculus.
First, we need to express the distance between a point on the parabola and the point (1, 4). A point on the parabola can be represented as (x, y). Since the parabola equation gives us a relationship between x and y, we can write y = ±√(2x). The distance D between the point (x, y) on the parabola and the point (1, 4) can be expressed as:
D = √((x – 1)² + (y – 4)²)
To simplify our work, we can minimize the square of the distance D² instead, since minimizing D² also minimizes D:
D² = (x – 1)² + (y – 4)²
Substituting y² = 2x into the distance formula, we rewrite y² as 2x.
Next, to minimize D², we need to find the critical points. We express D² in terms of x only:
D² = (x – 1)² + (√(2x) – 4)²
Now, expand and simplify this expression:
D² = (x – 1)² + (2x – 8√(2x) + 16)
Take the derivative of D² with respect to x and set it to zero to find critical points:
Through solving the derivative equation, we obtain a critical point x = 0.5 (after simplification).
Using the parabola equation y² = 2(0.5) = 1, we find:
y = ±1
Thus, the points on the parabola that are closest to (1, 4) are (0.5, 1) and (0.5, -1).
Therefore, the point on the parabola y² = 2x that is closest to the point (1, 4) will be either (0.5, 1) or (0.5, -1) depending on whether we are looking for the upper or lower point.