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r cos2θ = a

∵ In polar form,

radius of curvature = ρ 

= \(\frac{(r^2+(\frac{dr}{dθ})^2)^{3/2}}{r^2+2(\frac{dr}{dθ})^2-r\frac{d^2r}{dθ^2}}\) .........(1)

∵ r cos2θ = a

⇒ r = a sec2θ

⇒ \(\frac{dr}{dθ}\) = 2a sec2θ tan2θ

⇒ \(\frac{d^2r}{dθ^2}\) = 4a sec2θ tan²2θ + 4a sec³2θ

          = 4a sec2θ (sec²2θ +  tan²2θ)

From (1),

∴ ρ = \(\frac{(a^2sec^22θ + 4a^2sec^22θ\,tan^22θ)^{3/2}}{a^2sec^22θ+8a^2\,sec^22θ\,tan^22θ-4a^2\,sec^22θ(sec^22θ+tan^22θ)}\)

= \(\frac{a^3\,sec^32θ(1+4\,tan^22θ)^{3/2}}{a^2\,sec^22θ(1+8\,tan^22θ-4sec^22θ-4\,tan^22θ)}\)

= \(\frac{a\,sec2θ(1+4\,tan^22θ)^{3/2}}{1+4\,(tan^22θ-sec^22θ)}\)

= \(\frac{a\,sec2θ\,(1+4\,tan^22θ)^{3/2}}{1-4}\)

(∵ sec²θ - tan²θ = 1)

= \(-\frac{1}{3}\)a sec2θ (1+4 tan²2θ)^3/2