Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• In section 13.22 we proved that a conic symmetric about the origin satisfies the equation $\|X - F\| = \left|eX \cdot N - a\right|,$ where $a = ed + eF\cdot N.$ Use this relation to prove that $\|X - F\| + \|X + F\| = 2a$ if the conic is an ellipse. In other words, the sum of the distances from any point on an ellipse to its foci is constant.
• From properties of the dot product (and by extension, the norm) we know that $\|-X - F\| = \|X + F\|.$ Use this in conjunction with equation (13.29). $$\|X - F\| = e\left|(X - F) \cdot N - d\right| = \left|X \cdot N - F \cdot N - d\right| = \left|eX \cdot N - a\right|$$ Recall also that for a conic symmetric about the origin, $$a = \frac{ed}{1 - e^{2}}; \quad F = \frac{e^{2}d}{1 - e^{2}}N$$
• We know that if a conic is symmetric about the origin, then for every $X$ on the curve, $-X$ must also be on the curve. Thus, if $X$ satisfies the equation $\|X - F\| = \left|eX \cdot N - a\right|,$ then $-X$ satisfies the equation $$\|-X - F\| = \left|-eX \cdot N - a\right|$$But we know from the properties of the norm that $\|-X - F\| = \|X + F\|,$ thus $$\|X - F\| + \|X + F\| = \left|eX \cdot N - a\right| + \left|-eX \cdot N - a\right|$$But since the conic is symmetric about the origin, we can set $a = \frac{ed}{1 - e^{2}}$ turning the above equation into $$\|X - F\| + \|X + F\| = \left|eX \cdot N - \frac{ed}{1 - e^{2}}\right| + \left|-eX \cdot N - \frac{ed}{1 - e^{2}}\right|$$Since the equation $\|X - F\| = \left|eX \cdot N - a\right|$ presumes that $X$ is in the negative half-plane, then we assume also that $X$ is to the right of the origin. That is, $eX \cdot N > 0.$ But this implies that $$\left|eX \cdot N - a\right| = a - eX \cdot N$$because the horizontal distance from the origin to X must be smaller than the horizontal distance from the origin to the right directrix. In other words, $eX \cdot N < e\frac{d}{1-e^{2}}$ (given that $e < 1$ and $N = \textbf{i}$). And if we assume that $X$ is to the right of the origin, then $\left|-eX \cdot N - a\right|$ becomes $eX \cdot N + a$ since we know that $\frac{ed}{1 - e^{2}} > 0$ because $d > 0$ and $e < 1.$ Taking the sum of these terms, we see that $$\displaylines{\|X - F\| + \|X + F\| = a - eX \cdot N + eX \cdot N + a\\=2a\quad \blacksquare}$$