Mathematical Immaturity

14.9 Exercises

• A line perpendicular to the tangent line of a plane curve is called a normal line. If the normal line and a vertical line are drawn at any point of a certain plane curve $C,$ they cut off a segment of length 2 on the x-axis. Find a Cartesian equation for this curve if it passes through the point $(1,2).$ Two solutions are possible.
• The normal line has slope $m = -\frac{dx}{dy}.$ If the normal line for a given point $(x, y)$ always moves $y$ units vertically for every two units moved horizontally (left or right), this implies that the normal line through $(x, y)$ has one of two point-slope forms: $$ \begin{align*} (y - 0) = m[x - (x - 2)] \quad \text{or} \quad (y - 0) = m[x - (x + 2)] \end{align*} $$ Use this with $m = -\frac{dx}{dy}$ to find the relation between position and slope.
• By definition, the slope of the normal line must be $-\frac{dx}{dy}.$ And since we know that the change in vertical position is $y$ units for every two units of horizontal position changed (left or right), this means that the normal line is described by one of two possible point-slope equations $($since the normal line could intercept the $x-$axis at either $x + 2$ or $x - 2):$ $$ \begin{align*} (y - 0) = m[x - (x - 2)] \quad \text{or} \quad (y - 0) = m[x - (x + 2)] \end{align*} $$ Plugging in our value for $m$ and simplifying terms, we get $$ \begin{align*} y = 2\frac{dx}{dy} \quad \text{or} \quad y = -2\frac{dx}{dy} \end{align*} $$ As we can see, these are both separable differential equations. Separating terms and integrating, we get: $$ \begin{align*} \frac{1}{2}\int y\,dy &= \int dx \quad \text{or} \quad -\frac{1}{2}\int y\,dy = \int dx\\ \end{align*} $$ $$ \begin{align*} \frac{y^2}{4} - x = C_0 \quad \text{or} \quad \frac{y^2}{4} + x = C_1 \end{align*} $$ Plugging in our initial value $(x, y) = (1, 2),$ we find that $C_0 = 0$ and $C_1 = 2,$ giving us the two Cartesian equations: $$ y^2 = 4x \quad \text{or} \quad y^2 + 4x = 8 $$