Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Show that the tangent line to the ellipse $x^2/a^2 + y^2/b^2 = 1$ at the point $(x_0, y_0)$ has the equation $x x_0/a^2 + y y_0/b^2 = 1.$
• Recall from Exercise 10 that the vector $N$ normal to the ellipse at $(x_0, y_0)$ is \begin{align*} \\ N &= \frac{x_0}{a^2}\mathbf{i} + \frac{y_0}{b^2}\mathbf{j} \end{align*} Then, recall from Section 13.4 that the Cartesian equation of a line can be expressed in terms of a point $P$ on the line and the normal vector $N$ by $(X - P) \cdot N = 0.$
• Let $P = (x_0, y_0)$ be a point on the ellipse. We know from Exercise 10 that the vector $N$ normal to the ellipse at $P$ is \begin{align*} \\ N &= \frac{x_0}{a^2}\mathbf{i} + \frac{y_0}{b^2}\mathbf{j} \end{align*} Then, if $X = (x, y)$ is an arbitrary point on the line tangent to the ellipse at $P,$ we can express the Cartesian equation of the line in terms of $X,$ $P,$ and $N$ by the dot product relation $(X - P) \cdot N = 0.$ In other words, \begin{align*} \\ \frac{x_0}{a^2}(x - x_0) + \frac{y_0}{b^2}(y - y_0) &= 0 \end{align*} Rearranging terms gives us \begin{align*} \\ \frac{x}{a^2}x_0 + \frac{y}{b^2}y_0 &= \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} \end{align*} And since we know that $(x_0, y_0)$ is on the ellipse, $x_0^2/a^2 + y_0^2/b^2 = 1,$ giving us \begin{align*} \\ \frac{x}{a^2}x_0 + \frac{y}{b^2}y_0 &= 1 \quad \blacksquare \end{align*}