- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.21 Miscellaneous Review Exercises
- Show that the tangent line to the hyperbola $x^2/a^2 - y^2/b^2 = 1$ at the point $(x_0, y_0)$ is given by the equation $x_0x/a^2 - y_0y/b^2 = 1.$
- Use the same approach as in Exercise 11.
- From Exercise 16, we know that the vector normal to the hyperbola at point $P = (x_0, y_0)$ is given by \begin{align*} \\ N &= \frac{x_0}{a^2}\mathbf{i} - \frac{y_0}{b^2}\mathbf{j} \end{align*} The Cartesian equation for the line containing point $P$ with normal vector $N$ can be expressed as $(X - P)\cdot N = 0.$ In other words, if $X = (x, y),$ we have: \begin{align*} \\ \frac{x_0}{a^2}(x - x_0) - \frac{y_0}{b^2}(y - y_0) &= 0 \end{align*} Expanding and rearranging terms, we get \begin{align*} \\ \frac{x_0}{a^2}x - \frac{y_0}{b^2}y &= \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \end{align*} But we know that $P$ is a point on the hyperbola, which means $x_0^2/a^2 - y_0^2/b^2 = 1,$ giving us: \begin{align*} \\ \frac{x_0}{a^2}x - \frac{y_0}{b^2}y &= 1 \quad \blacksquare \end{align*}