[Not to be turned in]
In class, I showed you that that hyperboloid of one sheet is a "ruled surface", which means that it can be generated by a collection of straight lines.
Consider the hyperboloid given by $x^2+y^2z^2 = 1$. Recall that when $z=0$, $x^2+y^2=1$, so the points on the unit circle in the $xy$plane are part of this circle
 Show that the intersection of this surface with the plane $y=1$ is a pair of crossing lines.
 Note that the plane $y=1$ intersects the $xy$plane in the line $y=1$, which is tangent to the unit circle, and the plane $y=1$ is the vertical plane over that line.
Argue that by rotational symmetry, the plane over any tangent line to the unit circle intersects the hyperboloid in a pair of lines. (These form the two symmetric rulings of the hyperboloid that I showed in class.)
Later, when we do more with images of functions, we will be able to show this explicitly fairly easily, but for now, here is a rather long way to go about it.
 For a point $(a,b)$ on the unit circle (i.e., $a$ and $b$ where $a^2+b^2=1$), so that the tangent line to the circle in the $xy$plane has equation $y={1ax\over b}$. (Hint, recall that for a circle, the radius is perpendicular to the tangent line. Then think about the slope of the radius from the origin to $(a,b)$, and use that to find the slope of the tangent line. use the pointslope formula to get the equation of the tangent line.)
 The point in the plane over the tangent line are the ones where $y={1ax\over b}$ and $z$ is any value. Plug this $y$ into the equation for the hyperboloid to find a relationship between $x$ and $z$ within that plane. Show that this relationship produces two lines (two linear functions $z = m_1x + b_1$ and $z=m_2x + b_2$). These represent the two ruling lines through the point $(a,b,0)$ on the hyperboloid.

