Question

In Exercise 40 through 43, consider the problem of fitting a conic through$\mathit{m}$given points${\mathit{P}}_{\mathbf{1}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}\mathbf{,}\mathbf{.}\mathbf{......}\mathbf{,}{\mathit{P}}_{\mathbf{m}}\mathbf{(}{\mathbf{x}}_{\mathbf{m}}\mathbf{,}{\mathbf{y}}_{\mathbf{m}}\mathbf{)}$in the plane; see Exercise 53 through 62 in section 1.2. Recall that a conic is a curve in${\mathit{ℝ}}^{\mathbf{2}}$that can be described by an equation of the form , $\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}\mathit{y}\mathbf{+}{\mathit{c}}_{\mathbf{4}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}_{\mathbf{5}}\mathit{x}\mathit{y}\mathbf{+}{\mathit{c}}_{\mathbf{6}}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$where at least one of the coefficients is non zero.

40. Explain why fitting a conic through the points ${\mathit{P}}_{\mathbf{1}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}\mathbf{,}\mathbf{.}\mathbf{......}\mathbf{,}{\mathit{P}}_{\mathbf{m}}\mathbf{(}{\mathbf{x}}_{\mathbf{m}}\mathbf{,}{\mathbf{y}}_{\mathbf{m}}\mathbf{)}$amounts to finding the kernel of an$\mathit{m}\mathbf{\times }\mathbf{6}$matrix$\mathit{A}$. Give the entries of the row of $\mathit{A}$.

Note that a one-dimensional subspace of the kernel of defines a unique conic, since the equations$\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{0}$and$\mathit{k}\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{0}$describe the same conic.

Short answer

Thus, the entries for$ith$ row are$\left[\begin{array}{cccccc}1& x_i& y_i& x_i^2& x_i{y}_i& y_i^2\end{array}\right]$ .

Step-by-step solution

Given information 

A conic is a curve in${ℝ}^2$ that can be described by an equation of the form , $f(x,y)=c_1+c_{2}x+c_{3}y+c_4{x}^2+c_{5}xy+c_6{y}^2=0$where at least one of the coefficients $c_i$is non-zero.

Step 2:Explanation

To fit a conic through the points$P_1(x_1,y_1),\mathrm{.....},P_m(x_m,y_m)$ is equivalent to find the kernel of an $m{x}_6$of matrix and also these are solution to the equation.

$f(x,y)=c_1+c_{2}x+c_{3}y+c_4{x}^2+c_{5}xy+c_6{y}^2=0$

Thus, the$ith$ row have entries as follows:

$\left[\begin{array}{cccccc}1& x_i& y_i& x_i^2& x_i{y}_i& y_i^2\end{array}\right]$

Hence, the entries for$ith$ row are$\left[\begin{array}{cccccc}1& x_i& y_i& x_i^2& x_i{y}_i& y_i^2\end{array}\right]$ .