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Distance from a point on a parabola to the origin

Minimize the square of a distance

Substitute the curve equation into $d = x^{2} + y^{2}$ . Since square root increases with its input, the same $x$ minimizes $d$ and $d^{2}$ . Squaring often leaves a polynomial that is easier to optimize.

Problem

A point $P = (x, y)$ lies on $y = x^{2} - 9$ . Express its distance $d$ from the origin as a function of $x$ . For which positive $x$ is the distance smallest, and what is that distance?

Answer

d (x) = x^{4} - 17 x^{2} + 81

; the minimum occurs at

x = 17/2 \approx 2.92

, with

d = 35 /2 \approx 2.96

Checked by completing the square in d²; there is a matching minimum at negative x.

Step-by-step solution

1. Substitute the curve equation

From the distance formula, $d = x^{2} + y^{2}$ . With $y = x^{2} - 9$ ,

d (x) = x^{2} + (x^{2} - 9)^{2} = x^{4} - 17 x^{2} + 81 .

2. Minimize the expression under the root

Complete the square in $x^{2}$ :

d^{2} = (x^{2} - \frac{17}{2})^{2} + \frac{35}{4} .

The square cannot be negative, so the minimum occurs when $x^{2} = 17/2$ . For positive $x$ , this gives $x = 17/2$ .

3. Take the square root

d_{m i n} = 35/4 = \frac{35}{2} .

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