Find the curves for which R = N^3, where R is the radius of…

Find the curves for which $R = N^3,$ where $R$ is the radius of curvature and $N$ is the length of the normal (the segment of the normal between the point on the curve and the $x$-axis).

Step 1: Formulas and the governing equation

For $y=y(x)$, $R=\frac{(1+y'^2)^{3/2}}{|y''|}, \qquad N = y\sqrt{1+y'^2}.$

The condition $R=N^3$ reads $\dfrac{(1+y'^2)^{3/2}}{|y''|} = y^3(1+y'^2)^{3/2}$; dividing both sides by $(1+y'^2)^{3/2}\neq0$ leaves

\[ y'' = \pm\frac{1}{y^3}. \tag{1} \]

Step 2: Reduce the order — $y'=p(y)$

Since $x$ is absent, set $y'=p$, $y''=p\,dp/dy$. Equation (1) becomes $p\,dp=\pm dy/y^3$; integrating ($2C=a^2$), \[p^2 = a^2\mp\frac{1}{y^2} = \frac{a^2y^2\mp1}{y^2}, \qquad \frac{dy}{dx} = \pm\frac{\sqrt{a^2y^2\mp1}}{y},\] where the top sign throughout corresponds to $y''=+1/y^3$ and the bottom sign to $y''=-1/y^3$.

Step 3: Second integration

Separating variables and integrating the left side with $u=a^2y^2\mp1$: \[\frac{y\,dy}{\sqrt{a^2y^2\mp1}}=\pm dx \qquad\Longrightarrow\qquad \frac{\sqrt{a^2y^2\mp1}}{a^2}=\pm(x-b).\]

Squaring gives $a^2y^2 = a^4(x-b)^2\pm1$.

Step 4: The two families

Step 5: General solution

With $C_1=a^2>0$ and $C_2=b$, the two families combine into one two-parameter answer:

Each member is a hyperbola centred on the $x$-axis at $(C_2,0)$: sign $+$ gives hyperbolas with the $y$-direction transverse axis (from $y''=1/y^3$), sign $-$ gives hyperbolas with the $x$-direction transverse axis (from $y''=-1/y^3$).

Step 6: Verification

Take $y^2=a^2(x-b)^2+1/a^2$ (sign $+$). Differentiating, $2yy'=2a^2(x-b)$, so $y'=a^2(x-b)/y$, and

\[ y'' = \frac{a^2y-a^2(x-b)y'}{y^2} = \frac{a^2y^2-a^4(x-b)^2}{y^3} = \frac{1}{y^3}, \]

since $a^2y^2-a^4(x-b)^2=1$ by the equation of the curve itself — confirming $y''=1/y^3$, consistent with this being the $+$ case. Now check $R=N^3$ for $a=1,\,b=0$ (so $y^2=x^2+1$, $y'=x/y$, $y''=1/y^3$):

\[ R = \frac{\left(1+x^2/y^2\right)^{3/2}}{1/y^3} = (y^2+x^2)^{3/2}, \qquad N = y\sqrt{1+y'^2} = \sqrt{y^2+x^2}, \] so $N^3=(y^2+x^2)^{3/2}=R$. $\checkmark$