Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 2 — Applications of Ordinary First-Ordered Differential Equations

Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.

Solution:

Plot points on the curve,

A(x1,y1)A(x_{1},y_{1})

We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.

y−y1=−1m(x−x1)y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)

As the normal intersects the x-axis, y = 0

Substituting to the previous equation, we get

−y1=−1m(x−x1)−my1=−1(x−x1)−my1=−x+x1x=x1+my1\begin{align*} -y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\ -my_{1}&=-1\left(x-x_{1}\right)\\ -my_{1}&=-x+x_{1}\\ x&=x_{1}+my_{1} \end{align*}

By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1)

(x1−0)2+(y1−0)2=(x1+my1−x1)2+(0−y1)2x12+y12=m2y12+y12x12+y12=m2y12+y12x12=m2y12x1=m1y1    ;m=dydxx1=dydxy1\begin{align*} \sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\ \sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\ x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\ x_{1}^2&=m^2y_{1}^2\\ x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\ x_{1}&=\frac{dy}{dx}y_{1} \end{align*}

Change x1 and y1 to x and y,

x=ydydxxdx=ydy\begin{align*} x&=y\frac{dy}{dx}\\ xdx&=ydy \end{align*}

By integrating,

∫ ydx=∫ xdyy22=x22+Cy2=x2+2C\begin{align*} \int \:ydx&=\int \:xdy\\ \frac{y^2}{2}&=\frac{x^2}{2}+C\\ y^2&={x^2}+2C\\ \end{align*}

We get,

y2−x2−=2Cy^2-x^2-=2C
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