Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• The normal line at a point $P$ of a curve intersects the $x$-axis at $X$ and the $y$-axis at $Y.$ Find the curve if each $P$ is the mid-point of the corresponding line segment $XY$ and if the point $(4, 5)$ is on the curve.
• 1. Let $P = (x, y),$ $X = (x_0, 0),$ and $Y = (0, y_0).$ If $P$ is the midpoint between $X$ and $Y,$ what does that imply about the norms $\|P - X\|$ and $\|P - Y\|?$ What does this imply about $x_0$ and $y_0?$ 2. Let $P$ be parameterized by $x.$ What is its derivative? If $P,$ $X,$ and $Y$ are all on the line normal to the curve at $P,$ what does that imply about the line segments $(P-X)$ and $(P - Y)?$
• Let $X = (x_0, 0)$ $Y = (0, y_0),$ and $P = (x, y).$ Then, if $P$ is at the midpoint of the line segment $XY,$ by definition, $\|P - X\| = \|P - Y\|.$ Equivalently, this means that $\|P - X\|^2 = \|P - Y\|^2,$ or $(x - x_0)^2 + y^2 = x^2 + (y - y_0)^2.$ Simplifying terms, we get $x_0(x_0 - 2x) = y_0(y_0 - 2y),$ which is satisfied when $x_0 = 2x$ and $y_0 = 2y.$ To find the family of curves satisfying this condition, we note that $X$ and $P$ are both points on the line normal to the curve at $P.$ In other words, if $P$ is parameterized by $x,$ then the line segment $(P - X)$ is perpendicular to $P' = (1, \frac{dy}{dx}).$ That is, $(P - X)\cdot P' = 0,$ or \begin{align*} -x + y\frac{dy}{dx} &= 0 \end{align*} As we can see, this is a separable differential equation. Separating variables and integrating, we get the following family of hyperbolas \begin{align*} \frac{1}{2}y^2 &= \frac{1}{2}x^2 + C \end{align*} To find the specific curve containing $P = (4, 5),$ we simply set $x = 4$ and $y = 5$ and solve for $C.$ \begin{align*} \frac{1}{2}(25)&= \frac{1}{2}(16) + C \end{align*} As we can see, $C = 9/2,$ which means that the curve satisfying the initial conditions is the hyperbola $y^2 - x^2 = 9. \quad\blacksquare$