Question

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

55.$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)},\left(3,3\right)$.

Short answer

Thus, the cubic that passes through the nine given points is of the form .

$-2c_{10}x2C_{10}y+3c_{10}{x}^2-3c_{10}{y}^2-c_{10}{y}^3=0$

Step-by-step solution

Given in the question.

Each point $P_i\left(x_i,y_i\right)$defines an equation in the 10 variables given by:

${\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{2}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{3}}\mathbf{}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{4}}\mathbf{}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{5}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_6\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{7}}\mathbf{+}{\mathbf{x}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{8}}\mathbf{}\mathbf{+}{\mathbf{x}}_{\mathbf{i}}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{9}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{10}}\mathbf{=}\mathbf{0}$

There are ten points.

The system of ten equations is written as follows:

role="math" localid="1660371146567" $$\begin{gathered} \mathrm{A}\stackrel{\rightharpoonup }{\mathrm{c}}=\stackrel{\rightharpoonup }{0} \\ \mathrm{Here},\mathrm{A}=\left[\begin{array}{cccccccccc}1& {\mathrm{x}}_1& {\mathrm{y}}_1& {\mathrm{x}}_1^2& {\mathrm{x}}_1{\mathrm{y}}_1& {\mathrm{y}}_1^2& {\mathrm{x}}_1^3& {\mathrm{x}}_1^2{\mathrm{y}}_1& {\mathrm{x}}_1{\mathrm{y}}_1^2& {\mathrm{y}}_1^3\\ 1& {\mathrm{x}}_2& {\mathrm{y}}_2& {\mathrm{x}}_2^2& {\mathrm{x}}_2{\mathrm{y}}_2& {\mathrm{y}}_2^2& {\mathrm{x}}_2^3& {\mathrm{x}}_2^2{\mathrm{y}}_2& {\mathrm{x}}_2{\mathrm{y}}_2^2& {\mathrm{y}}_2^3\\ 1& {\mathrm{x}}_3& {\mathrm{y}}_3& {\mathrm{x}}_3^2& {\mathrm{x}}_3{\mathrm{y}}_3& {\mathrm{y}}_3^2& {\mathrm{x}}_3^3& {\mathrm{x}}_3^2{\mathrm{y}}_3& {\mathrm{x}}_3{\mathrm{y}}_3^2& {\mathrm{y}}_3^3\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& {\mathrm{x}}_9& {\mathrm{y}}_9& {\mathrm{x}}_9^2& {\mathrm{x}}_9{\mathrm{y}}_9& {\mathrm{y}}_9^2& {\mathrm{x}}_9^3& {\mathrm{x}}_9^2{\mathrm{y}}_9& {\mathrm{x}}_9{\mathrm{y}}_9^2& {\mathrm{y}}_9^3\end{array}\right] \end{gathered}$$

Apply gauss-Jordan elimination in the matrix .

Plug in the ten points to derive the matrix.

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

Now, use gauss-Jordan elimination to solve the system $A\stackrel{\rightharpoonup }{c}=\stackrel{\rightharpoonup }{0}$. Note that the A matrix is identical to the A matrix from Exercise 54, with the addition of one row. Thus, the first nine rows is replaced with row echelon form in Exercise 54.

$$\begin{gathered} \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 2& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right] \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right] \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right] \end{gathered}$$

Showing that cubics through (0,0)(1,0)(2,0)(0,1)(1,1)(2,1)(0,2)(1,2)(2,2)(3,3).

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

$$\begin{gathered} c_1=0 \\ {c}_2=2c_{10} \\ {c}_3=2c_{10} \\ {c}_4=3c_{10} \\ {\mathrm{c}}_5=0 \\ {\mathrm{c}}_6=-3c_{10} \\ {\mathrm{c}}_7=-c_{10} \\ {\mathrm{c}}_8=0 \\ {\mathrm{c}}_9=0 \end{gathered}$$

While $c_{10}$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 .

$$\begin{gathered} -2c_{10}x+2c_{10}y+3c_{10}{x}^2-3c_{10}{y}^2+c_{10}{y}^3+c_{10}{y}^3=0 \\ {c}_{10}\left(2x-3x^2-x^3+2y-3y^2+y^3\right)=0 \\ y\left(y-1\right)\left(y-2\right)-x\left(x-1\right)\left(x-2\right)=0 \end{gathered}$$

Sketch of cubics.

As the above equation is satisfied if x = y.It is also satisfied if ( x , y ) lies on a certain ellipse centred at (1,1). Consider the equation of an ellipse at (1,1).

$e\left(x,y\right)={\left(x-1\right)}^2+{\left(y-1\right)}^2+2b\left(x-1\right)\left(y-1\right)-c=0$

Define the function as:

$g\left(x,y\right)=\left(x-y\right)\left[{\left(x-1\right)}^2+{\left(y-1\right)}^2+2b\left(x-1\right)\left(y-1\right)-c\right]$

Thenifor, ie ifis on the line of slope one through the origin, or is on the ellipse.

Expand the formula for:

$$\begin{gathered} g\left(x,y\right)=\left(x-y\right)\left[x^2-2\left(b+1\right)x+y^2-2\left(b+1\right)y+2\left(b+1\right)-c\right] \\ =x^3-2\left(b+1\right)x^2+\left[2\left(b+1\right)-c\right]x-y^3+2\left(b+1\right)y^2-\left[2\left(b+1\right)-c\right]y \end{gathered}$$

Now, let$b=\frac{1}{2}$and c=1, then it gives:

$$\begin{gathered} g\left(x,y\right)=x^3-3x^2+2x-y^3+3y^2-3y \\ =x\left(x-1\right)\left(x-2\right)-y\left(y-1\right)\left(y-2\right) \end{gathered}$$

Therefore, the function g (x,y) constructed here is equal to the cubic that passes through the given points. Thus, the cubic contains the line x = y and the ellipse e (x,y).

The cubic is shown as follows: