Question

Differential equation for all straighy lines which are at unit distance from origin

Answer 1

Answer:

I'm going to assume that the OP really just wants the general equation for all lines for which the minimum distance to the origin is 1. A general equation for a straight line is given by:

$[math]y=mx+b[/math]$

Where $[math]m[/math]$is the slope, and $[math]b[/math$is the $[math]y[/math]$intercept of the line. We want to find the coordinates of the point on this line for which the distance to the origin is a minimum. We will take advantage of the fact that when the distance is minimized, the distance squared is also minimized, in order to keep the math cleaner. Let the distance to the origin be [math]D[/math], then we have:

$[math]D^2=(x-0)^2+(y-0)^2=x^2+(mx+b)^2[/math]$

We can find the [math]x[/math] coordinate that minimize this distance by differentiating with respect to [math]x[/math], and setting this equal to zero:

$[math]\frac{dD^2}{dx}=2x+2m(mx+b)=0\implies x+m^2x+mb=(1+m^2)x+mb=0[/math]$

$[math]\implies x=\frac{-mb}{1+m^2}[/math]$

Now, when the distance is one, the distance squared is also one, so we can plug this value of [math]x[/math] into the [math]D^2[/math] formula, and set it equal to one to find how [math]b[/math] and [math]m[/math] must be related:

$[math]D^2=x^2+(mx+b)^2=(\frac{-mb}{1+m^2})^2+(m(\frac{-mb}{1+m^2})+b)^2=1[/math]$

$[math]\implies \frac{m^2b^2}{(m^2+1)^2}+(\frac{-m^2b+m^2b+b}{m^2+1})^2=1[/math]$

$[math]\implies \frac{m^2b^2}{(m^2+1)^2}+(\frac{b}{m^2+1})^2=1[/math]$

$[math]\implies \frac{m^2b^2}{(m^2+1)^2}+\frac{b^2}{(m^2+1)^2}=1[/math]$

$[math]\implies \frac{m^2b^2+b^2}{(m^2+1)^2}=1[/math]$

$[math]\implies \frac{b^2(m^2+1)}{(m^2+1)^2}=1[/math]$

$[math]\implies \frac{b^2}{m^2+1}=1[/math]$

$[math]\implies b^2=m^2+1[/math]$

$[math]\implies b=\pm\sqrt{m^2+1}[/math]$

Plugging this into the general equation for a line, we arrive at the general equation for all lines that have a minimum distance to the origin of one:

$[math]y=mx\pm\sqrt{m^2+1}[/math] for all [math]m[/math]$