91影视

Each point $P_i(x_i,y_i)$defines an equation in the 10 variables$c_1,c_2,\mathrm{.....},c_{10}$ given by:

$c_1+x_i{c}_2+y_i{c}_3+x_i^2{c}_4+x_i{y}_i{c}_5+y_i^2{c}_6+x_i^3{c}_7+x_i^2{y}_i{c}_8+x_i{y}_i^2{c}_9+y_i^3{c}_{10}=0$,

There are eight points.

The system of five equations is written as follows:

$A\stackrel{鈫�}{c}=\stackrel{鈫�}{0}$

Where$A=\left[\begin{array}{cccccccccc}1& x_1& y_1& x_1^2& x_1{y}_1& y_1^2& x_1^3& x_1^2{y}_1& x_1{y}_1^2& y_1^3\\ 1& x_2& y_2& x_2^2& x_2{y}_2& y_2^2& x_2^3& x_2^2{y}_2& x_2{y}_2^2& y_2^3\\ 1& x_3& y_3& x_3^2& x_3{y}_3& y_3^2& x_3^3& x_3^2{y}_3& x_3{y}_3^2& y_3^3\\ 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�\\ 1& x_8& y_8& x_8^2& x_8{y}_8& y_8^2& x_8^3& x_8^2{y}_8& x_8{y}_8^2& y_8^3\end{array}\right]$.

Plug in the eight points to derive the$A$matrix.

$A=\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 3& 0& 9& 0& 0& 27& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 0& 3& 0& 0& 9& 0& 0& 0& 27\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\end{array}\right]$

Now, use gauss-Jordan elimination to solve the system $A\stackrel{鈫�}{c}=\stackrel{鈫�}{0}$.

The solution of the equation$A\stackrel{鈫�}{c}=\stackrel{鈫�}{0}$ which satisfies:

$\begin{array}{l}{c}_1=0\\ c_2=0\\ c_3=2c_{10}\\ c_4=0\\ c_5=鈭�c_8鈭�c_9+c_{10}\\ c_6=鈭�3c_{10}\\ c_7=0\\ c_{10}=0\end{array}$

While$c_8,c_9$ are free variables. Recall that the cubic equation is as follows:

$c_1+x{c}_2+y{c}_3+x^2{c}_4+xy{c}_5+y^2{c}_6+x^3{c}_7+x^{2}y{c}_8+x{y}^2{c}_9+y^3{c}_{10}=0$

Therefore, the cubic that passes through the eight given points is of the form,

$\begin{array}{c}(鈭�c_8鈭�c_9)xy+c_8{x}^{2}y+c_{9}x{y}^2=0\\ c_8(x^{2}y鈭�xy)+c_9(x^{2}y鈭�xy)=0\\ c_{8}xy(x鈭�1)+c_{9}xy(y鈭�1)=0\\ xy(c_8(x鈭�1)+c_9(y鈭�1))=0\end{array}$

As the first example, substitute . $c_8=0,c_9=1$The cubic is$xy(x鈭�1)=0$.

Now, for a point $(x,y)$on the cubic curve is either $x=0,y=0\text{鈥�}or\text{鈥�}y=1$. This set is graphed as follows:

As the first example, substitute . $c_8=1,c_9=0$The cubic is$xy(y鈭�1)=0$.

Now, for a point $(x,y)$on the cubic curve is either$x=0,x=1\text{鈥�}or\text{鈥�}y=0$. This set is graphed as follows: