Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Prove that the product of the perpendicular distances from an arbitrary point on a hyperbola to its asymptotes is constant.
• For a hyperbola with standard-form equation: \begin{align*} \\ \frac{x^2}{a^2} - \frac{y^2}{b^2} &= 1 \end{align*} its asymptotes are the lines $y = -bx/|a|$ and $y = bx/|a|.$ For $P = (x, y)$ on the hyperbola, the perpendicular distance $d$ is given by \begin{align*} d &= \frac{|(P - X) \cdot N|}{\|N\|} \end{align*} where $X$ is a point on the asymptote and $N$ is the vector normal to the asymptote.
• Suppose we have a hyperbola with standard-form equation: \begin{align*} \\ \frac{x^2}{a^2} - \frac{y^2}{b^2} &= 1 \end{align*} Where $b = |a|\sqrt{e^2 - 1},$ $a = ed/(1 - e^2),$ and $d$ is the distance from the focus to the directrix. From Section 13.23, we know that the asymptotes of this hyperbola are the lines $y = -bx/|a|$ and $y = bx/|a|.$ Let $P = (x, y)$ be an arbitrary point on the hyperbola. Note that the asymptote lines $y_1 = -bx/|a|$ and $y_2 = bx/|a|$ pass through the origin. As such, we can express the perpendicular distance from $P$ to the lines given by $y_1$ and $y_2$ as $d_1 = |P \cdot N_1|/\|N_1\|$ and $d_2 = |P \cdot N_2|/\|N_2\|,$ where \begin{align*} \\ N_1 &= \frac{b}{|a|}\mathbf{i} + \mathbf{j}, \quad N_2 = \frac{b}{|a|}\mathbf{i} - \mathbf{j} \end{align*} are the vectors normal to the asymptote lines $y_1$ and $y_2,$ respectively. The product of the two distances is then \begin{align*} \\ d_1d_2 &= \frac{|P \cdot N_1||P \cdot N_2|}{\|N_1\|\|N_2\|} \\ \\ &= \frac{\left|bx/|a| + y\right|\left|bx/|a| - y\right|}{(b^2/a^2) + 1} \\ \\ &= \frac{b^2x^2 - a^2y^2}{b^2 + a^2} \end{align*} But we know that all points $P$ on the hyperbola satisfy $x^2/a^2 - y^2/b^2 = 1,$ so we can rewrite $b^2x^2 - a^2y^2 = a^2b^2$ to give us \begin{align*} \\ d_1d_2 &= \frac{a^2b^2}{a^2 + b^2} \end{align*} which is constant for all $P$ on the hyperbola. $\,\blacksquare$