Find point where radius of curvature is minimum

Question

Find the point where radius of curvature is minimum for the curve

$$x^2y=a\left(x^2+\frac{a^2}{\sqrt{5}}\right)$$

Answer 1

Couldn't you simply rewrite the equation as $$y=a\left(1+\frac{a^2}{x^2\sqrt{5}}\right)$$ and differentiate? I get $$y'(x)=-\frac{2a^3}{x^3\sqrt{5}}$$ and $$y''(x)=\frac{6a^3}{x^4\sqrt{5}}.$$

With this knowledge, you have a formula for the radius of curvature and a method of checking for extrema of a singular variable function. Do you think you can take it from here?