Answer
Result See proof
Work Step by Step
Step 1 a) Recall the way we've defined the eccentricity of the ellipse. By using the notation from the book, the eccentricity is given as a ratio \[ \epsilon = \frac{a}{c} \] and we know that in the ellipse the distance from the focus to the closest point is: \[ r_0 = a - c \] while the distance from the focus to the farthest point is \[ r_1 = a + c \] But now the claim is the direct result of these two definitions. Step 2 b) Notice that the values \(r_{\text{max}}\) and \(r_{\text{min}}\) are simply extreme values of \[ \| \mathbf{r} \|^2 \] The square does not matter since the quadratic function is monotonous so we simply square the length function to get rid of the square root. But since these values are extremes then the following relations have to hold: \[ \begin{align*} \| \mathbf{r} \|^2 &= \| \mathbf{r} \cdot \mathbf{r} \| \\ &= 2 \mathbf{r} \cdot \mathbf{r}' \\ &= 0 \end{align*} \] Since \(\mathbf{r}' = \mathbf{v}\) it follows that \(\mathbf{v}\) and \(\mathbf{r}\) are orthogonal. Step 3 c) We proceed in the same way as in part b). \[ \begin{align*} \| \mathbf{v}_{\text{min}} \|^2 &\text{ and } \| \mathbf{v}_{\text{max}} \|^2 \text{ are extremes of } \| \mathbf{v} \|^2 \\ &\text{From here:} \\ \| \mathbf{v} \|^2 &= 2\mathbf{v} \cdot \mathbf{v}' = 0 \end{align*} \] since \(\mathbf{a} = \mathbf{v}'\) it follows that \(\mathbf{v}\) and \(\mathbf{a}\) are orthogonal. Step 4 d) This part of the task requires a bit more elaborate analysis. Notice that \(\mathbf{r} \times \mathbf{v} = \mathbf{b}\). So it follows that: \[ \| \mathbf{r} \| = \| \mathbf{r} \times \mathbf{v} \| = \| \mathbf{r} \| \cdot \| \mathbf{v} \| \cdot \sin \theta = \| \mathbf{b} \| = \| \mathbf{r} \times \mathbf{v} \| \] Now, the discussion turns into discussing the right side of the equality. since the angle is \(\frac{\pi}{2}\), the value of sine function is \(1\). Due to the fact that \(\mathbf{b}\) has to be a constant vector, the maximum of \(\mathbf{r}\) has to occur at the minimum of \(\mathbf{v}\) and vice versa. From here it follows that: \[ \| \mathbf{r} \| = \| \mathbf{r} \times \mathbf{v} \| = r_{\text{max}} \cdot v_{\text{min}} = r_{\text{min}} \cdot v_{\text{max}} \]
