Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Prove that a point $(x_0, y_0)$ is inside, on, or outside the ellipse $x^2/a^2 + y^2/b^2 = 1$ according as $x_0^2/a^2 + y_0^2/b^2$ is less than, equal to, or greater than 1.
• Recall Theorem 13.19, Equation 13.34, and Equation 13.35: Theorem 13.19. Let $C$ be a conic section with eccentricity $e \neq 1$ and with a focus $F$ at a distance $d$ from a directrix $L.$ If $N$ is a unit normal to $L$ and if $F = eaN,$ where $a = ed/(1 - e^2),$ then $C$ is the set of all points $X$ satisfying the equation \begin{align*} \\ \|X\|^2 + e^2a^2 &= e^2\left(X \cdot N\right)^2 + a^2 \quad (13.34) \end{align*} Then, putting $X$ into rectangular coordinates, [ie, $X = (x, y)$] and setting $N = \mathbf{i}$ (meaning the directrix is vertical), Equation (13.34) becomes $x^2 + y^2 + e^2a^2 = e^2x^2 + a^2,$ or $x^2(1 - e^2) + y^2 = a^2(1 - e^2),$ giving us \begin{align*} \\ \frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} &= 1 \quad (13.35) \end{align*}
• If $X = (x_0, y_0)$ satisfies \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \end{align*} with $b^2 = a^2(1 - e^2),\ 0 < e < 1,$ we know from Theorem 13.19 that $X$ is on the conic $C$ with focus $F = eaN,$ where $a = ed/(1 - e^2),$ $d$ is the distance from $F$ to the vertical directrix line $L$ located at $x = a/e,$ and $N$ is the unit vector normal to $L.$ Now, suppose $X$ is such that \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} &> 1 \end{align*} Then, $x^2(1 - e^2) + y^2 > a^2(1 - e^2).$ In other words, \begin{align*} \|X\|^2 + \|F\|^2 > e^2\left(X \cdot N\right)^2 + a^2 \end{align*} Subtracting $2(F \cdot X)$ from both sides, we get \begin{align*} \|X - F\|^2 > (ex - a)^2 \end{align*} But the directrix of $C$ is located at $x = a/e,$ so we can write $(ex - a)^2$ as $e^2(x - \frac{a}{e})^2,$ which gives us: \begin{align*} \\ \|X - F\|^2 > e^2d(X, L)^2 \end{align*} Where $d(X, L)$ is the distance from $X$ to the directrix $L.$ But this is equivalent to the inequality $\|X - F\| > ed(X, L),$ which implies that $X$ lies closer to the directrix than it does to the focus. In other words, $X$ lies outside of the ellipse $C.$ Now, if $X$ is such that: \begin{align*} \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} &< 1 \end{align*} we can use a similar argument to show that $\|X - F\| < ed(X, L)$ and hence $X$ lies inside the ellipse $C.$ This completes the proof.