[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 $x^2+y^2-z^2 = 1$. 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 $y={1-ax\over b}$. (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 point-slope formula to get the equation of the tangent line.)
- The point in the plane over the tangent line are the ones where $y={1-ax\over b}$ 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.
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