Show that the hyperboloid of one sheet is a doubly ruled surface.

Question

In "Vector Calculus" by Michael Corral, Chapter 1, section 6, there is a question, that is:

11. Show that the hyperboloid of one sheet is a doubly ruled surface, i.e. each point on the surface is on two lines lying entirely on the surface (Hint: Write equation (1.35) as $\frac{x^2}{a^2}-\frac{z^2}{c^2} = 1 - \frac{y^2}{b^2}$, factor each side. Recall that two planes intersect in a line.

The question has its hint, telling you to rewrite the equation (1.35), which is the equation of a hyperboloid of one sheet: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$$, then factor each side. Doing this leads to:

$$\frac{x^2}{a^2}-\frac{z^2}{c^2} = 1 - \frac{y^2}{b^2}$$ then, $$\frac{(cx)^2-(az)^2}{(ac)^2}=\frac{b^2 - y^2}{b^2}$$ and then $$\frac{(cx - az)(cx + az)}{(ac)^2}=\frac{(b-y)(b+y)}{b^2}$$

From here, I'm stuck. I don't know what to do with this equation. Looking back, the second hint suggest that we refer to the fact that two planes intersect in a line. That line of intersection, L, between two planes, is defined as: $$L: \textbf{r} +t(\textbf{n}_1\times\textbf{n}_2) $$ for $-\infty < t < \infty$, $\textbf{n}_1$ is the normal vector of the first plane $P_1$, and $\textbf{n}_2$ is the normal vector of the second plane $P_2$.

I can't think of any ideas to connect this two together, maybe because I haven't thought enough, but still, I can't find any way to solve it on my own. Please suggest how to prove this problems (not to show all the steps, but if you are, it's a pleasure), and tell me where I'm wrong.

(Maybe this topic has already been discussed, so far I can't find anything much related, so possibly it may be a duplicate)

Answer 1

Hint : Try the factorization $$\left(\frac{x}{a}-\frac{z}{c}\right)\left(\frac{x}{a}+\frac{z}{c}\right)=\left(1-\frac{y}{b}\right)\left(1+\frac{y}{b}\right)$$

See if you can extract 2 pairs of planes from this. If so, then both the pairs would give you a straight line (which lie entirely on the surface since that's how you crafted the planes), which would intersect at the required point $(x,y,z)$.

Edit (Some more hints) : From the factorization, you can see that $$\frac{\frac{x}{a}-\frac{z}{c}}{1-\frac{y}{b}}=\frac{1+\frac{y}{b}}{\frac{x}{a}+\frac{z}{c}}=\lambda_1\ (\text{say})$$

Given $(x_0,y_0,z_0)$ on the hyperboloid, you can easily find out $\lambda_1$.

Then a pair of planes is given by $$\left(\frac{x}{a}-\frac{z}{c}\right)=\lambda_1\left(1-\frac{y}{b}\right)$$ $$\lambda_1\left(\frac{x}{a}+\frac{z}{c}\right)=\left(1+\frac{y}{b}\right)$$

You can check that the line given by the solution of these 2 planes is indeed on the hyperboloid.

The other pair of planes is given by taking the ratio the other way around.

Answer 2

Let $(x,y,z)$ be any point of the hyperboloid with one sheet satisfying the equation

$$S(x,y,z) = \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}-1=0.$$ We have

$$\left(\frac{x}{a} - \frac{z}{c}\right)\left(\frac{x}{a} + \frac{z}{c}\right)=\left(1 - \frac{y}{b}\right)\left(1 + \frac{y}{b}\right) $$ For $(x_0,y_0,z_0)$ a given point of $S$, we can define four planes of equations

$$P(\epsilon_1,\epsilon_2, x_0) \equiv \left(\frac{x}{a} +\epsilon_1 \frac{z}{c}\right)\left(1 +\epsilon_2 \frac{y_0}{b}\right) - \left(\frac{x_0}{a} +\epsilon_1 \frac{z_0}{c}\right)\left(1 +\epsilon_2 \frac{y}{b}\right)=0$$ where $\epsilon_1, \epsilon_2 \in \{-1,1\}$.

From the equation above, which is equivalent to the definition of our surface $S(x,y,z)=0$, You can verify that any point of $S$ belonging to the intersection of $P(1,1,x_0)$ and $P(-1,-1,x_0)$ belongs to $S$. Similarly for any point of $S$ belonging to the intersection of $P(1,-1,x_0)$ and $P(-1,1,,x_0)$.

From this you get the desired conclusion.