Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• The normal line at each point of a curve and the line from that point to the origin form an isosceles triangle whose base is on the $x$-axis. Show that the curve is a hyperbola.
• 1. If the point $P$ on the $x$-axis forms an isosceles triangle with the origin and $X = (x, y),$ what does this imply about the norm of $P - X?$ 2. If $P$ and $X$ are both points on the line normal to the curve at $X,$ what is the Cartesian equation of the normal line? Parameterize $y$ by $x$ and take its derivative to find the slope of the line tangent to $X.$
• Let $X = (x, y)$ be a point on the curve, parameterized by $x.$ Its derivative is $X' = (1, \frac{dy}{dx}).$ The line normal to the curve at $X$ is then parallel to $N = (\frac{dy}{dx}, -1).$ Now, let $P = (x_0, 0)$ be a point at the intersection of the $x$-axis and normal line such that the line segments $OX$ and $PX$ form an isosceles triangle whose base of length $x_0$ is on the $x$-axis. In other words, we have $\|P - X\| = \|X\|,$ or \begin{align*} \\ (x_0 - x)^2 + y^2 &= x^2 + y^2 \end{align*} We can see from this relation that $x_0^2 = 2x_0x.$ Assuming that the triangle has a nonzero base, we can see that $x_0 = 2x.$ To find $y$ in terms of its parameter $x,$ we note that since $X$ and $P$ are both points on the normal line, $(P - X)$ is parallel to $N$ and perpendicular to $X'.$ Thus, the Cartesian equation of the normal line is $(P - X) \cdot X' = 0$ or: \begin{align*} \\ x - y\frac{dy}{dx} &= 0 \end{align*} But as we can see, this is a separable differential equation. Separating variables and integrating, we get \begin{align*} \\ \frac{x^2}{2} &= \frac{y^2}{2} + C \end{align*} where $C$ is the sum of the constants of integration. But as we can see, by rearranging variables, we get $x^2 - y^2 = 2C$ which is the equation for a hyperbola. $\,\blacksquare$