Find the coordinates of the point(s) on the parabola y = 4 - x^2 that is closest to the point (0,...

Question:

Find the coordinates of the point(s) on the parabola {eq}y = 4 - x^2 {/eq} that is closest to the point {eq}(0, \; 1) {/eq}.

Minimizing distance:

The distance between any curve to the point {eq}(a, \; b, ) {/eq} is:

{eq}d= \sqrt{ (x-a)^2 + (y - b)^2 } {/eq}

Next, we neeed to compute the value of {eq}x {/eq} and {eq}y {/eq} in order to get the desired solution.

Answer and Explanation:

The distance between the curve {eq}y = 4 - x^2 {/eq} at the point {eq}(0,1) {/eq} is:

{eq}d= \sqrt{ (x-0)^2 + (y - 1)^2 } = \sqrt{ x^2 + (y-1)^2...

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from PSAT Prep: Tutoring Solution

Chapter 10 / Lesson 13
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