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Point-Quadratic Distance

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PointQuadraticCurve

To find the minimum distance between a point in the plane (x_0,y_0) and a quadratic plane curve

y=a_0+a_1x+a_2x^2,
(1)

note that the square of the distance is

r^2=(x-x_0)^2+(y-y_0)^2
(2)
=(x-x_0)^2+(a_0+a_1x+a_2x^2-y_0)^2.
(3)
PointQuadraticDistance

Minimizing the distance squared is equivalent to minimizing the distance (since r^2 and |r| have minima at the same point), so take

(partial(r^2))/(partialx)=2(x-x_0)+2(a_0+a_1x+a_2x^2-y_0)(a_1+2a_2x)=0.
(4)

Expanding,

x-x_0+a_0a_1+a_1^2x+a_1a_2x^2-a_1y_0+2a_0a_2x+2a_1a_2x^2+2a_2^2x^3-2a_2y_0x=0
(5)
2a_2^2x^3+3a_1a_2x^2+(a_1^2+2a_0a_2-2a_2y_0+1)x+(a_0a_1-a_1y_0-x_0)=0.
(6)

Minimizing the distance to find the closest point (x^*,y^*) therefore requires solution of a cubic equation.

See also

Point-Line Distance--2-Dimensional

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Cite this as:

Weisstein, Eric W. "Point-Quadratic Distance." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/Point-QuadraticDistance.html

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