About another method of finding the radius of curvature, given by Norman J. Wildberger (a person which sometimes seems to be a "crank"):
The radius of a circle due to an inscribed triangle $\triangle ABC$ with sides $a,b,c$ is $R=\dfrac{abc}{4S_\triangle}$ .
He uses this to find the radius of curvature formula of $y=x^2$ .
But he uses the cross product for the area: $$\begin{align} &A(p,p^2)\ ,\ B(q,q^2)\ ,\ C(r,r^2)\\\\ a^2&=(p-q)^2\bigl[1+(p+q)^2\bigr]\\b^2&=(q-r)^2\bigl[1+(q+r)^2\bigr]\\c^2&=(r-p)^2\bigl[1+(r+p)^2\bigr]\\\\ \vec A-\vec C&=\binom{p-r}{p^2-r^2}\ ,\ \vec B-\vec C=\binom{q-r}{q^2-r^2}\\\\ S_\triangle&=\frac{\Big|(\vec A-\vec C)\times(\vec B-\vec C)\Big|}{2}\\ &=\frac{\Big|(p-r)(q^2-r^2)-(p^2-r^2)(q-r)\Big|}{2}\\ &=\frac{\Big|(p-r)(q-r)(q+r)-(p-r)(p+r)(q-r)\Big|}{2}\\ &=\frac{\Big|(p-q)(q-r)(r-p)\Big|}{2}\end{align}$$ Now $$\begin{align} \lim_{p,q,r\to t}r&=\lim_{p,q,r\to t}\frac{\sqrt{(p-q)^2(q-r)^2(r-p)^2\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{4\dfrac{\bigl|(p-q)(q-r)(r-p)\bigr|}{2}}\\\\ &=\lim_{p,q,r\to t}\frac{\bigl|(p-q)(q-r)(r-p)\bigr|\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{2\bigl|(p-q)(q-r)(r-p)\bigr|}\\\\ &=\lim_{p,q,r\to t}\frac{\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{2}=\frac{\sqrt{(1+4t^2)^3}}{2} \end{align}$$ But when I tried using Heron's formula for the area and plugging it in, this did not work and I did not get any cancellation terms: $$R=\frac{abc}{4\dfrac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}}$$ Can anyone tell me why?
$ \begin{align} &A(p,p^2)\ ,\ B(q,q^2)\ ,\ C(r,r^2)\\\\ a^2&=(p-q)^2\bigl[1+(p+q)^2\bigr]\\b^2&=(q-r)^2\bigl[1+(q+r)^2\bigr]\\c^2&=(r-p)^2\bigl[1+(r+p)^2\bigr]\\\\ \end{align} $
I used this version to substitute in the denominator
$a^2= \left(p^2-q^2\right)^2+(p-q)^2,\\b^2= \left(q^2-r^2\right)^2+(q-r)^2,\\c^2= \left(r^2-p^2\right)^2+(r-p)^2$
$$R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{abc}{\sqrt{-a^4 + 2 a^2 b^2 - b^4 + 2 a^2 c^2 + 2 b^2 c^2 - c^4}}$$
$$R=\frac{|p-q||q-r||r-p|\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{\sqrt{den}}$$
where $den = 2 \left(\left(p^2-q^2\right)^2+(p-q)^2\right) \left(\left(r^2-p^2\right)^2+(r-p)^2+\left(q^2-r^2\right)^2+(q-r)^2\right)-\left(-\left(r^2-p^2\right)^2-(r-p)^2+\left(q^2-r^2\right)^2+(q-r)^2\right)^2-\left(\left(p^2-q^2\right)^2+(p-q)^2\right)^2\quad(*)$
$$R=\frac{|p-q||q-r||r-p|\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{\sqrt{4 (p-q)^2 (p-r)^2 (q-r)^2}}$$
$$R=\frac{|p-q||q-r||r-p|\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{2 |p-q| |p-r||q-r|}$$
$$R=\frac{\sqrt{\bigl[1+(p+q)^2\bigr]\bigl[1+(q+r)^2\bigr]\bigl[1+(r+p)^2\bigr]}}{2}$$
and the limit obviously gives the same result
$(*)$ the factorization was made with the help of Mathematica
Hope this can help