Question

Check whether or not the conicoid represented by
5x^2+4y^2-4yz+2xz+2x-4y-8z+2=0 is central or not. If it is, transform the
equation by shifting the origin to the centre. Else, change any one coefficient to make the equation that of a central conicoid.

Answer 1

Answer on Question #78592 – Math – Analytic Geometry

Question

Check whether or not the conicoid represented by

$$5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0$$

is central or not. If it is, transform the equation by shifting the origin to the centre. Else, change any one coefficient to make the equation that of a central conicoid.

Solution

Theorem 1

The origin $O(0,0,0)$ is a centre of the conicoid

$$F(x,y,z) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0$$

if and only if $u = v = w = 0$

We have that

$$u = 1 \neq 0, v = -2 \neq 0, w = -4 \neq 0$$

Therefore, the origin $O(0,0,0)$ is not the centre of the conicoid.

Theorem 2

A conicoid $S$, given by the equation

$$F(x,y,z) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0$$

has the point $P(x_0,y_0,z_0)$ as a centre if and only if

$$\left\{ \begin{array}{l} ax_0 + hy_0 + gz_0 + u = 0 \\ hx_0 + by_0 + fz_0 + v = 0 \\ gx_0 + fy_0 + cz_0 + w = 0 \end{array} \right.$$

We have that

$$\left\{ \begin{array}{l} 5x_0 + z_0 + 1 = 0 \\ 4y_0 - 2z_0 - 2 = 0 \\ x_0 - 2y_0 - 4 = 0 \end{array} \right.$$

Augmented matrix

$$\begin{pmatrix} 5 & 0 & 1 & -1 \\ 0 & 4 & -2 & 2 \\ 1 & -2 & 0 & 4 \end{pmatrix}$$ $$\begin{pmatrix} 5 & 0 & 1 & -1 \\ 0 & 4 & -2 & 2 \\ 1 & -2 & 0 & 4 \end{pmatrix} \xrightarrow{R_1/5} \begin{pmatrix} 1 & 0 & 1/5 & -1/5 \\ 0 & 4 & -2 & 2 \\ 1 & -2 & 0 & 4 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 0 & 1/5 & -1/5 \\ 0 & 4 & -2 & 2 \\ 1 & -2 & 0 & 4 \end{pmatrix} \xrightarrow{R_3-R_1} \begin{pmatrix} 1 & 0 & 1/5 & -1/5 \\ 0 & 4 & -2 & 2 \\ 0 & -2 & -1/5 & 21/5 \end{pmatrix}$$ $$\begin{array}{l} \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 4 & -2 & 2 \\ 0 & -2 & -1/5 & 21/5 \end{array} \right) \xrightarrow{R_2/4} \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & -2 & -1/5 & 21/5 \end{array} \right) \\ \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & -2 & -1/5 & 21/5 \end{array} \right) \xrightarrow{R_3+(2)R_2} \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & -6/5 & 26/5 \end{array} \right) \\ \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & -6/5 & 26/5 \end{array} \right) \xrightarrow{(-5/6)R_3} \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & -13/3 \end{array} \right) \\ \left( \begin{array}{cccc} 1 & 0 & 1/5 & -1/5 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & -13/3 \end{array} \right) \xrightarrow{R_1-(\frac{1}{5})R_3} \left( \begin{array}{cccc} 1 & 0 & 0 & 2/3 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & -13/3 \end{array} \right) \\ \left( \begin{array}{cccc} 1 & 0 & 0 & 2/3 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & -13/3 \end{array} \right) \xrightarrow{R_2+(\frac{1}{2})R_3} \left( \begin{array}{cccc} 1 & 0 & 0 & 2/3 \\ 0 & 1 & 0 & -5/3 \\ 0 & 0 & 1 & -13/3 \end{array} \right) \\ \left\{ \begin{array}{l} x_0 = \frac{2}{3} \\ y_0 = -\frac{5}{3} \\ z_0 = -\frac{13}{3} \end{array} \right. \\ \end{array}$$

Then the point $P\left(\frac{2}{3}, -\frac{5}{3}, -\frac{13}{3}\right)$ is the centre of the coincoid.

In order for the coincoid to become central, it is necessary to apply a shift of the form

$$\left\{ \begin{array}{l} x^* = x - \frac{2}{3} \\ y^* = y + \frac{5}{3} \\ z^* = z + \frac{13}{3} \end{array} \right.$$

Then

$$5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0$$ $$5 \left(x^* + \frac{2}{3}\right)^2 + 4 \left(y^* - \frac{5}{3}\right)^2 - 4 \left(y^* - \frac{5}{3}\right) \left(z^* - \frac{13}{3}\right) + 2 \left(x^* + \frac{2}{3}\right) \left(z^* - \frac{13}{3}\right) + 2 \left(x^* + \frac{2}{3}\right) \left(2x^2 - \frac{13}{3}\right) + 4$$ $$+ 2 \left(x^* + \frac{2}{3}\right) - 4 \left(y^* - \frac{5}{3}\right) - 8 \left(z^* - \frac{13}{3}\right) + 2 = 0$$ $$5x^{*2} + \frac{20}{3}x^* + \frac{20}{9} + 4y^{*2} - \frac{40}{3}y^* + \frac{100}{9} - 4y^*z^* + \frac{52}{3}y^* + \frac{20}{3}z^* - \frac{260}{9} +$$ $$\begin{array}{l} +2x^*z^* - \frac{26}{3}x^* + \frac{4}{3}z^* - \frac{52}{9} + 2x^* + \frac{4}{3} - 4y^* + \frac{20}{3} - 8z^* + \frac{104}{3} + 2 = 0 \\ 5x^{*2} + 4y^{*2} - 4y^*z^* + 2x^*z^* + 0 \cdot x^* - 0 \cdot y^* + 0 \cdot z^* + \frac{70}{3} = 0 \\ \end{array}$$

Hence

$$\begin{array}{l} 5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 \rightarrow \\ \rightarrow 5x^{*2} + 4y^{*2} - 4y^*z^* + 2x^*z^* + \frac{70}{3} = 0 \\ \end{array}$$

**Answer:**

1) The conicoid represented by

$$5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0$$

is not central.

2) The point $P\left(\frac{2}{3}, -\frac{5}{3}, -\frac{13}{3}\right)$ is the centre of the coincoid.

3) It is necessary to apply a shift of the form

$$\left\{ \begin{array}{l} x^* = x - \frac{2}{3} \\ y^* = y + \frac{5}{3} \\ z^* = z + \frac{13}{3} \\ \end{array} \right.$$

to make the conicoid central.

$$\begin{array}{l} 5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 \rightarrow \\ \rightarrow 5x^{*2} + 4y^{*2} - 4y^*z^* + 2x^*z^* + \frac{70}{3} = 0 \\ \end{array}$$

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