How do you find the points on the parabola $y = 6 - x^{2}$ $y = 6 - x^2$ that are closest to the point (0,3)?

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How do you find the points on the parabola $y = 6 - x^{2}$ $y = 6 - x^2$ that are closest to the point (0,3)?

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Answer 1 · 2015-08-16T20:56:39

Substitute the equation of the parabola into the distance formula to get the square root of a quartic to minimise. This is quadratic in $x^{2}$ $x^2$ , so complete the square to find the minimum.

Explanation:

If $(x, y)$ $(x, y)$ is a point on the parabola, then the distance between $(x, y)$ $(x, y)$ and $(0, 3)$ $(0, 3)$ is:

$\sqrt{{(x - 0)}^{2} + {(y - 3)}^{2}}$ $sqrt((x-0)^2+(y-3)^2)$

$= \sqrt{x^{2} + {(6 - x^{2} - 3)}^{2}}$ $= sqrt(x^2+(6-x^2-3)^2)$

$= \sqrt{x^{2} + {(3 - x^{2})}^{2}}$ $=sqrt(x^2+(3-x^2)^2)$

$= \sqrt{x^{2} + 9 - 6 x^{2} + x^{4}}$ $=sqrt(x^2+9-6x^2+x^4)$

$= \sqrt{x^{4} - 5 x^{2} + 9}$ $=sqrt(x^4-5x^2+9)$

$= \sqrt{{(x^{2} - \frac{5}{2})}^{2} + \frac{11}{4}}$ $=sqrt((x^2-5/2)^2+11/4)$

This will have its minimum value when ${(x^{2} - \frac{5}{2})}^{2} = 0$ $(x^2-5/2)^2 = 0$ , that is when $x = \pm \sqrt{\frac{5}{2}} = \pm \frac{\sqrt{10}}{2}$ $x = +-sqrt(5/2) = +-sqrt(10)/2$

When $x = \pm \frac{\sqrt{10}}{2}$ $x = +-sqrt(10)/2$ we have $y = 6 - x^{2} = 6 - \frac{5}{2} = \frac{7}{2}$ $y = 6 - x^2 = 6 - 5/2 = 7/2$

So the points on the parabola at minimum distance from $(0, 3)$ $(0, 3)$ are:

$(\pm \frac{\sqrt{10}}{2}, \frac{7}{2})$ $(+-sqrt(10)/2, 7/2)$

As a bonus, we have also calculated the minimum distance as:

$\sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$ $sqrt(11/4) = sqrt(11)/2$

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How do you find the points on the parabola $y = 6 - x^{2}$ $y = 6 - x^2$ that are closest to the point (0,3)?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the points on the parabola y=6−x2y = 6 - x^2 that are closest to the point (0,3)?

1 Answer

Explanation:

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Impact of this question

How do you find the points on the parabola $y = 6 - x^{2}$ $y = 6 - x^2$ that are closest to the point (0,3)?