Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Given an ellipse $x^2/a^2 + y^2/b^2 = 1.$ Show that the vectors $T$ and $N$ given by \begin{align*} \\ T = -\frac{y}{b^2}\,\mathbf{i} + \frac{x}{a^2}\,\mathbf{j}, \quad N = \frac{x}{a^2}\,\mathbf{i} + \frac{y}{b^2}\,\mathbf{j} \end{align*} are, respectively, tangent and normal to the ellipse when placed at the point $(x,y).$ If the eccentric angle of $(x_0, y_0)$ is $\theta_0,$ show that the tangent line at $(x_0, y_0)$ has the Cartesian equation \begin{align*} \ \frac{x}{a} \cos \theta_0 + \frac{y}{b} \sin \theta_0 = 1. \end{align*}
• To find the vector tangent to the ellipse, first parameterize the coordinates $(x,y)$ in terms of $x.$ The derivative of this vector will be parallel to the tangent vector. To find the relation between the derivative vector and $dy/dx,$ differentiate the ellipse's standard form equation with respect to $x.$ To show that the tangent line at $(x_0, y_0)$ has Cartesian equation \begin{align*} \\ \frac{x}{a} \cos \theta_0 + \frac{y}{b} \sin \theta_0 = 1. \end{align*} recall Example 3 of Section 14.6 which gives the relation between a point $(x, y)$ on an ellipse and its eccentic angle $\theta.$ 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 any point $(x, y)$ on the ellipse be parameterized by $x.$ Then, the vector $T$ tangent to the ellipse at $(x, y)$ is parallel to $\mathbf{i} + \frac{dy}{dx}\mathbf{j}.$ To find $\frac{dy}{dx},$ we take the derivative of the standard form equation with respect to $x,$ giving us \begin{align*} \\ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} &= 0 \end{align*} This means that $dy/dx$ satisfies the relation \begin{align*} \\ -\frac{y}{b^2}\frac{dy}{dx} &= \frac{x}{a^2} \end{align*} But as we can see, multiplying the vector $\mathbf{i} + \frac{dy}{dx}\mathbf{j}$ by the scalar $-y/b^2$ gives us $T.$ This shows that $T$ is tangent to the ellipse at $(x,y).$ To show that $N$ is normal to the ellipse at $(x, y)$ we simply take the dot product $T \cdot N$ and note that it is zero. If $\theta_0$ is the eccentric angle of $(x_0, y_0),$ then the coordinates are related to the angle by the following equations: \begin{align*} \\ x_0 &= a\cos\theta_0, \quad y_0 = b\sin\theta_0 \end{align*} Using our previously derived normal vector $N,$ we can express the Cartesian equation of the line tangent to the ellipse at $(x_0, y_0)$ in terms of the dot product relation $(X - P) \cdot N = 0,$ where $X$ is a point $(x, y)$ on the line and $P = (x_0, y_0).$ This gives us: \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, the equation becomes: \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*} But since we know that $x_0 = a\cos\theta_0,$ $y_0 = b\sin\theta_0,$ and $x_0^2/a^2 + y_0^2/b^2 = 1,$ we get: \begin{align*} \\ \frac{x}{a}\cos\theta_0 + \frac{y}{b}\sin\theta_0 &= 1 \quad \blacksquare \end{align*}