Find the minimum distance between a point in the plane
and a quadratic Plane Curve
![\begin{displaymath}
y=a_0+a_1x+a_2x^2.
\end{displaymath}](p2_927.gif) |
(1) |
The square of the distance is
Minimizing the distance squared is the equivalent to minimizing the distance (since
and
have minima at the
same point), so take
![\begin{displaymath}
{\partial (r^2)\over \partial x} = 2(x-x_0)+2(a_0+a_1x+a_2x^2-y_0)(a_1+2a_2x)=0
\end{displaymath}](p2_933.gif) |
(3) |
![\begin{displaymath}
x-x_0+a_0a_1+{a_1}^2+a_1a_2x^2-a_1y_0+2a_0a_2x+2a_1a_2x^2+2{a_2}^2x^3-2a_2y_0x=0
\end{displaymath}](p2_934.gif) |
(4) |
![\begin{displaymath}
2{a_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.
\end{displaymath}](p2_935.gif) |
(5) |
Minimizing the distance therefore requires solution of a Cubic Equation.
See also Point-Point Distance--1-D, Point-Point Distance--2-D,
Point-Point Distance--3-D
© 1996-9 Eric W. Weisstein
1999-05-25