How can we find the radius of curvature for this equation $$(x^2+y^2)^2=a^2(x^2-y^2)$$ Here I know the formula but I just want to find y' and y'' I'm stuck in it
here is what know $$\delta=\frac{(1+y'^2)^\frac{3}{2}}{y''}$$
i got y' $$y'=\frac{x}{y}\frac{a-2\sqrt{(x^2-y^2)}}{a+2\sqrt{(x^2-y^2)}}$$
It is easier to do this in polar coordinates,
$(x^2+y^2)^2=a^2(x^2-y^2)$
Using $x = r \cos\theta, y = r \sin\theta$,
$r^4 = a^2 r^2 \cos2\theta \implies r^2 = a^2 \cos2\theta$
$r_{\theta} = - \cfrac{a^2 \sin2\theta}{r} \implies r_{\theta}^2 = \cfrac{a^4 - r^4}{r^2}$
$r_{\theta \theta} = \cfrac{a^2 \sin2\theta}{r^2} r_{\theta} - \cfrac{2 a^2 \cos2\theta}{r} = - \cfrac{r_{\theta}^2}{r} - 2 r$
Radius of curvature, $R = \cfrac{(r^2 + r_{\theta}^2)^{3/2}}{|r^2 + 2 r_{\theta}^2 - r r_{\theta \theta}| } = \cfrac{\sqrt{r^2 + r_{\theta}^2}}{3}$
$ = \cfrac{a^2}{3 r} = \cfrac{a^2}{3 \sqrt{x^2+y^2}}$