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5.6.5.1 Implicitization

We can eliminate $ v$ to form the resultant $ F(u)$ , then solve $ F(u)=0$ for $ u$ and use the inversion algorithm to obtain $ v$ .

Example 5.6.2. Let us consider an ellipse and a circle

$\displaystyle f = \frac{x^2}{4} + y^2 -1 =0\;,$      
$\displaystyle g = (x-1)^2 +y^2 - 1 =0\;,$      

as in Fig. 5.16.

Figure 5.16: Ellipse and circle intersection
\begin{figure}\centerline{ \psfig{figure=fig/ellipse_circ.eps,height=2in} }\end{figure}

First we eliminate $ y$ from these two equations. This leads to

$\displaystyle 3x^2-8x+4 = 0\;,$      

which has two real roots $ x = 2$ and $ x = \frac{2}{3}$ . These lead to $ y^2 = 0$ and $ y^2 = \frac{8}{9}$ , respectively.

However there are possible numerical problems at the tangential intersection point $ x=2,\; y=0$ . Let us assume that due to error $ \epsilon>0$ , we have

$\displaystyle x = 2+\varepsilon\;,$      

hence
$\displaystyle y^2 = - \varepsilon ( 1 + \frac{\varepsilon}{4}) < 0\;.$      

This implies that $ y$ is imaginary and that no real roots exist. This would have as a consequence missing an intersection solution, leading to a robustness problem.

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Next: 5.6.5.2 Newton's method Up: 5.6.5 Implicit algebraic/implicit algebraic Previous: 5.6.5 Implicit algebraic/implicit algebraic   Contents   Index
December 2009