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Point-Quadratic Distance

Find the minimum distance between a point in the plane $(x_0,y_0)$ and a quadratic Plane Curve

\end{displaymath} (1)

The square of the distance is
$\displaystyle r^2$ $\textstyle =$ $\displaystyle (x-x_0)^2+(y-y_0)^2$  
  $\textstyle =$ $\displaystyle (x-x_0)^2+(a_0+a_1x+a_2x^2-y_0)^2.$ (2)

Minimizing the distance squared is the equivalent to minimizing the distance (since $r^2$ and $\vert r\vert$ have minima at the same point), so take
{\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} (3)

\end{displaymath} (4)

\end{displaymath} (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