For Problem DD-7:

One end of the semiminor axis is the point where $r = {\frac 1 2}$, $s = 0$; here x$= R \left[\begin{array}{r}{\frac 1 2}\\ 0\end{array}\right] = [v _ 1 \vert v ... ... = \left[\begin{array}{r}\frac {\sqrt 2}4\\ \frac {\sqrt 2}4\end{array}\right]$. It can be checked that this point is on the curve.

Note: Sometimes ``semimajor axis'' means the length of the line segment from the origin and sometimes it means the line segment itself. We can tell by the context.

For Problem DD-8:

(a) As in the earlier explanation, when we start with x$^ t A$   x$= 1$ and substitute x$= R$   r we get r$^ t R ^ t A R$   r$= 1$, or r$^ t D$   r$= 1$, where $D = \left[\begin{array}{rr}\lambda _ 1&0\\ 0&\lambda _ 2\end{array}\right]$. Written out, r$^ t D$   r$= 1$ becomes $\lambda _ 1 r ^ 2 + \lambda _ 2 s ^ 2 = 1$.

(b) If $\lambda _ 1$ and $\lambda _ 2$ are both positive, we get an ellipse (or a circle, if they are equal). If one is positive and the other negative, we get a hyperbola. If one is positive and the other is 0, we get two parallel lines. In all other cases there are no points.

Note: It is interesting to consider taking $\lambda _ 1$ to be a fixed positive number and gradually changing $\lambda _ 2$, starting from $\lambda _ 1$ and then going negative. We'd see an ellipse that expands along its major axis until it becomes two parallel lines, which then bend back and becomes a hyperbola.

For Problem DD-9:

(a) Here $A = \left[\begin{array}{rr}6&2\\ 2&9\end{array}\right]$, as in Problem DD-. In that problem we found $\lambda _ 1 = 10$, $\lambda _ 2 = 5$, so the transformed equation is $10 r ^ 2 + 5 s ^ 2 = 1$, giving semiaxes $\frac 1{\sqrt 10}$ and $\frac 1{\sqrt 5}$. Since the first is smaller, it's the semiminor axis and the second is the semimajor axis.

(b) In the earlier problem we saw $R = \left[\begin{array}{rr}\frac 1{\sqrt 5}& -\frac 2{\sqrt 5}\\ \frac 2{\sqrt 5}& \frac 1{\sqrt 5}\end{array}\right]$ (called $P$ there). With x$= R$r, for $r = 1$, $s = 0$ on the $r$-axis we have x$= R \left[\begin{array}{r}1\\ 0\end{array}\right] = \left[\begin{array}{r}\frac 1{\sqrt 5}\\ \frac 2{\sqrt 5}\end{array}\right]$. The slope of the $r$ axis is the $y$-value over the $x$-value, so is 2.