- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
13.25. Miscellaneous review exercises on conic sections
- Refer to Exercise 11. Prove that on each branch of a hyperbola the difference $\|X - F\|$ - $\|X + F\|$ is constant.
- The setup is the same as in Exercise 11, but with the added change that for a hyperbola, the value $a = \frac{ed}{1 - e^{2}}$ is less than zero since $e > 1$ for a hyperbola
- Referring to the setup from Exercise 11, we know that $$\|X - F\| - \|X + F\| = \left|eX \cdot N - a\right| - \left|-eX \cdot N - a\right|$$If X is to the right of the origin, then $\left|eX \cdot N - a\right| = eX \cdot N - a$ since $a = \frac{ed}{1 - e^{2}},$ which is less than zero for a hyperbola. And since $\left|X \cdot N\right| < \left|\frac{d}{1 - e^{2}}\right|,$ $\left|-eX \cdot N - a\right| = -eX \cdot N - a.$ Combining these two results, we get $$\displaylines{\|X - F\| - \|X + F\| = eX \cdot N - a - eX \cdot N - a\\ =-2a \quad \blacksquare}$$