Generating lines Show that the generators of the hyperboloid $\frac{x^2}{25}+\frac{y^2}{16}–\frac{z^2}{4}=1$ which are parallel to the plane $4x-5y-10z + 7 = 0$ are $x+5=0, y+2z=0$ and $y+4=0, 2x=5z$
My attempt
Equation of generators of $\mu$ system of the given hyperboloid is
$$(\frac X5+\frac z2)= \mu(1+\frac y4), (\frac x5–\frac z2)=(\frac1\mu)(1-\frac y4)$$
This generator will be parallel to the given plane if $\frac{\frac 15}4 = \frac{-\frac\mu4}{–5} = \frac{\frac 12}{–10} = (\frac{-\mu}7) = k$ (say).
This implies that $1=20k$ and $-1=20k$ implies that $k=0$
Therefore $\mu=0$
Therefore $\frac x5+\frac z2=0,y=4$
But this is not the answer.
 A: I agree with your equations :
$$\begin{cases}(\frac x5+\frac z2)&=&\mu(1+\frac y4)\\(\frac x5–\frac z2)&=&(\frac1\mu)(1-\frac y4)\end{cases}\tag{1}$$
Indeed their product gives back the equation of the hyperboloid. But you must be conscious that (1) describes only one of the two families of generating lines, the other one being :
$$\begin{cases}(\frac y4+\frac z2)&=& \lambda(1+\frac x5)\\(\frac y4–\frac z2)&=&(\frac1\lambda)(1-\frac x5)\end{cases}\tag{2}$$
The main result we must know for solving your problem is that a directing vector of a line defined in this way (i.e., as the intersection of 2 planes):
$$\begin{cases}ax+by+cz&=&d\\a'x+b'y+c'z&=&d'\end{cases} \ \text{is given by cross product} \ \begin{pmatrix}a\\b\\c\end{pmatrix} \times \begin{pmatrix}a'\\b'\\c'\end{pmatrix}$$
In our case, we will have:
$$\begin{pmatrix}1/5\\-\mu/4\\1/2\end{pmatrix} \times \begin{pmatrix}1/5\\1/(4 \mu)\\-1/2\end{pmatrix}=\begin{pmatrix}\mu/8-1/(8 \mu)\\1/5\\1/(20 \mu)+\mu/20\end{pmatrix}$$
This vector must be orthogonal to the normal vector to the plane, a fact that we express as a dot product equal to $0$:
$$\color{red}{4}[\mu/8-1/(8 \mu)]+\color{red}{(-5)}(1/5)+\color{red}{(-10)}[1/(20 \mu)+\mu/20]=0$$
This equation will give a single value $\mu=-1$.
Inserting this value of $\mu$ into (1) gives :
$$\begin{cases}\frac x5+\frac z2+1+\frac y4&=&0\\\frac x5-\frac z2+1-\frac y4&=&0\end{cases}\tag{3}$$
This system of equations doesn't look to be any of the given ones. But by adding and subtracting equations (3), we obtain the equivalent system :
$$\begin{cases}x+5&=&0\\y+2z&=&0\end{cases}$$
which is one of the two equations that were given to you.
Where has the second one been "lost" ? Nowhere, we must just consider the other family described by (2) and do the same kind of (a little tedious) calculations.
