Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface.

Question

Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface generated by the movement of a line and explicite its rectilinear generatrices. I have tried to make its matrix, knowing that $a_{11}=1$, $a_{12}=3$...$a_{00}=8$.. But I don't think that's the way to solve it because the matrix that i will get will have a lot of 0's..

Answer 1

If you want to make a matrix do it like this.

$\mathbf x^T \begin {bmatrix} 1 &0 &0\\ 0 &3 &2\\ 0 &2 &0\\ \end{bmatrix}\mathbf x + \begin {bmatrix} -6&8&0\end{bmatrix} \mathbf x = 0$

Since that matrix is symmetric is is diagonalizable with ortho-normal basis.

$\mathbf x^t P^T D P \mathbf x + BP^TP\mathbf x = -8\\ \mathbf u = P \mathbf x$

$P = \begin {bmatrix} 1 &0 &0\\ 0 &\frac 2{\sqrt5} &-\frac 1{\sqrt5}\\ 0 &\frac 1{\sqrt5} &\frac 2{\sqrt5}\\ \end{bmatrix}$

$D = \begin {bmatrix} 1 &0 &0\\ 0 &4 &0\\ 0 &0 &-1\\ \end{bmatrix}$

$u_1^2 + 4u_2^2 - u_3^2 - 6u_1 + \frac {16}{\sqrt5} u_2 +\frac {8}{\sqrt5} u_3=0\\ (u_1-3)^2 + 4(u_2+\frac2{\sqrt 5})^2 - (u_3- \frac 4{\sqrt 5})^2 = 1 $

That is a hyperboliod of one sheet.

$(x-3)^2 + 4(\frac 2{\sqrt5} y - \frac 1{\sqrt5} z+\frac2{\sqrt 5})^2 - (\frac 1{\sqrt5} y + \frac 2{\sqrt5}z- \frac 4{\sqrt 5})^2 = 1 $

Answer 2

Here is a solution for the second part (explicitation of a family of generatrices)

Once you have the equation of surface $(S)$ under the form

$$\tag{1} (S) \ \ \ X^2+4Y^2-Z^2=0 \ \ \ \ \iff \ \ \ \ 4Y^2=Z^2-X^2 $$

(see solution by @Doug M), here is how one finds a system of generatrices.

Let us consider:

$$\tag{2}(G_{\lambda}) \ \begin{cases}2Y=\lambda(Z-X)\\2Y=\dfrac{1}{\lambda}(Z+X)\end{cases} \ \ \ \ \lambda \in \mathbb{R^*}$$

For each fixed value of $\lambda$, $(G_{\lambda})$ represents a line (Recall that in $\mathbb{R^3}$, a line has two equations, i.e., is represented as the intersection of 2 planes). Moreover :

$$\tag{3} (x,y) \ \text{verifies} \ (2) \ \implies \ (x,y) \ \text{verifies} \ (1)$$

(by equating the product of the LHSides and RHsides). What is the meaning of $(3)$ ? Plainly that

$$(G_{\lambda}) \subset (S).$$

Conversely, it is easy to prove that, if $(x,y) \in (S)$, there exist a unique $\lambda$ such that $(x,y) \in (G_{\lambda}).$

We have thus proved that $(S)$ is a ruled surface, "ruled" by generatrices $(G_{\lambda})$.

But that's not all the story; there is a second system of generatrices.

Let us permute in (2) the RHS, this time with a different parameter, $\mu$.

We get the family of lines:

$$(\Gamma_{\mu}) \ \begin{cases}2Y=\mu(Z+X)\\2Y=\dfrac{1}{\mu}(Z-X)\end{cases} \ \ \ \ \mu \in \mathbb{R^*}$$

which can as well define surface $(S)$ as a ruled surface.

This double way for the ruling of an hyperboloid with one sheet is nicely represented as an animation in (https://en.wikipedia.org/wiki/Hyperboloid)