Special condition the circle of curvature satisfies when its being plotted

Question

I waa trying to see what the nature of the circle of curvature is when i tried plotting for a point on a parabola

lets say we have a curve given by $y^2 = 4x$ ,take a point P on that curve , for making a the centre of circle of the curvature this point is making with the parabola curve we have a formula for how much the radius be involving first and second derivative . But suppose we want to find the centre we can first try making a tangent supoose we managed it , but then for that point and the tangent line the , there are infinite circles passing through that point and tangent to the tangent line of parabola.

• For the centre of curvature whats the unique feature i should look for when i try drawing many circles along that normal line of parabola ?

Is it that that unique circle some part (very close to that point) must be touching the parabola curve ? Other circles (infinite) will not have that feature or something else unique about it?
Suppose we get the propety to be something , how do we prove that its satisfied by only one circle at all .(i am not talking using the radius formula and say as radius is unique so there exists only one circle inside the curve . ). Like my question my raised when i tried to try plotting like this : many circles seems to be touching at that point P but still which oen is exactly the curvature one , i am still not getting : all seems to be satisfying that they are part of a circle and tangent to that point :

Answer 1

For a planar curve $\mathbf{r}(t) = (x(t) , y(t) ) $ , the circle having the same curvature has a center given by

$ C(t) = ( x(t) - \alpha y'(t) , y(t) + \alpha x'(t) ) $

with

$ \alpha = \dfrac{ x'^2 + y'^2 }{ x' y'' - y' x'' } $

And the radius of this circle is given by

$ R = \dfrac{ (x'^2 + y'^2 )^{3/2}} { x' y'' - y' x'' } $

For the curve $ y^2 = 4 x$ , the corresponding parametric equation is $\mathbf{r}(t) = ( t^2 , 2 t) $, therefore

$ (x', y') = (2 t , 2 ) $ and $ (x'', y'') = (2 , 0) $

Answer 2

At points $P$ and $P'$ of a parabola of focus $S$, construct lines $PN$, $PN'$ parallel to the axis of the parabola, and normals $PC$ (bisector of $\angle SPN=\alpha$), $P'C$ (bisector of $\angle SP'N'=\alpha'$). Let then $\phi=\angle PSP'$ and $\theta=\angle PCP'$.

From triangle $PVS$ (see figure below) we get $\alpha+\phi=\alpha'$, while from triangles $PSU$ and $P'CU$ we get $\alpha/2+\phi=\alpha'/2+\theta$. Comparing these equalities we thus find: $$ \phi=2\theta. $$ Consider now circle $s$ circumscribed to $SPP'$: its arc $\gamma_s=PP'$ subtended by $\phi$ has a length $\gamma_s=\phi\cdot d_s$, where $d_s$ is the diameter of $s$. And consider circle $c$ circumscribed to $CPP'$: its arc $\gamma_c=PP'$ subtended by $\theta$ has a length $\gamma_c=\theta\cdot d_c$, where $d_c$ is the diameter of $c$.

When $P'\to P$, $PC$ tends to the radius of curvature $\rho$ of the parabola at $P$, circle $c$ tends to the circle tangent to the parabola at $P$ having as diameter $\rho$, circle $s$ tends to the circle tangent to the parabola at $P$ and passing through $S$, having a diameter $\delta$ which can be easily constructed. But the ratio $\gamma_s/\gamma_c$ tends to $1$ as $P'\to P$, hence: $$ 1=\lim_{P\to P'}{\gamma_s\over\gamma_c}= \lim_{P\to P'}{\phi\cdot d_s\over\theta\cdot d_c}= \lim_{P\to P'}{2d_s\over d_c}={2\delta\over \rho}. $$ That is:

the radius of curvature at a point $P$ of a parabola is twice the diameter of the circle tangent to the parabola at $P$ and passing through the focus.

The same technique can be used to find the radius of curvature of an ellipse or hyperbola: see here for the details.