How to find the locus of intersection of normals Given a parabola $y^2 = x$, how can we find the equation of the curve formed by the intersection of normals drawn from different points?
I have attached an image for reference.

I tried taking two close points $x$ and $x+dx$ and tried finding the intersection of their normals and solving the differential equation to get the answer but I couldn't get it right. Could someone explain how to find this either using differential equations or by any other method.
 A: Assuming you are, as I asked above, after the curve formed by where each normal intersects its "immediate neighbor", let's expand a bit on this.
A different (perhaps more rigorous) way to frame the above is the following: At each point the parabola has a curvature, which is to say a best-fitting circle (something akin to a second derivative, but less bound to the orientation of the axes). This best-fitting circle has a center. We are after the curve that these centers follow as we move along the curve.
In that spirit, let's first parameterize the parabola as $\gamma(t)=(t^2,t)$. The issue is now, given a value $t$, how can we find the corresponding circle center?
First, we will find a unit normal vector. The velocity vector is given by $\gamma'(t)=(2t,1)$. A vector that is normal to this (and points in the right direction) is $(1,-2t)$. We normalize this to $\frac1{\sqrt{1+4t^2}}(1,-2t)$.
Next, we get the right length. It is well-known that for circular motion with constant speed $v$, radius $r$ and centripetal acceleration $a$ we have $v^2=ar$. We don't have constant speed, so we can't use $|\gamma''(t)|$ directly in place of $a$. We need the component of $\gamma''(t)$ which is normal to the curve (and thus normal to $\gamma'(t)$). But we already have a unit normal vector, so this is easy:
$$
a=\frac{1}{\sqrt{1+4t^2}}(1,-2t)\cdot(2,0)=\frac{2}{\sqrt{1+4t^2}}
$$
We can now find the radius:
$$
r=\frac{v^2}a=\frac{\gamma'(t)^2}{2/\sqrt{1+4t^2}}\\
=\frac{(1+4t^2)\sqrt{1+4t^2}}2
$$
Multiply this with our unit normal vector, and we have the vector from $\gamma(t)$ to the center. Add this to $\gamma(t)$, and we find the coordinates of the center:
$$
\gamma(t)+\frac{1}{\sqrt{1+4t^2}}(1,-2t)\cdot\frac{(1+4t^2)\sqrt{1+4t^2}}2\\
=\gamma(t)+\left(\frac{1+4t^2}2,-t(1+4t^2)\right)\\
=\left(\frac{1}2+3t^2,-4t^3\right)
$$
A: The idea is feasible, you might not have taken it to its conclusion.
The normal at $x$ is a line through $(x,f(x))$ with slope $(-f'(x),1)$ orthogonal to the tangent. Now to get where two of these lines intersect, you have to solve
\begin{align}
x_1-s_1f'(x_1)&=x_2-s_2f'(x_2)\\
f(x_1)+s_1 &= f(x_2)+s_2
\\[1em]\hline
\implies 
x_1+f'(x_2)f(x_1)-s_1(f'(x_1)-f'(x_2))&=x_2+f'(x_2)f(x_2)\\
s_1 &=\frac{x_2-x_1+f'(x_2)(f(x_2)-f(x_1))}{f'(x_2)-f'(x_1)}\\
\end{align}
Next, as $x_1\approx x_2$ are close together, one can replace the differences by derivatives according to the mean value theorem, or just form the difference quotients. Then in the limit $x_2\to x=x_1$
$$
s=\frac{1+f'(x)^2}{f''(x)}.
$$
