It's All Ratios

We just need the top portion of the bi-cone, then double the angle in the final result.

$V = \dfrac{1}{3}\pi r^2h = K$ and $S = \pi r\sqrt{r^2 + h^2}$ .

$V = \dfrac{1}{3}\pi r^2h = K \implies h = \dfrac{3K}{\pi r^2} \implies S(r) = \dfrac{\sqrt{\pi^2 r^6 + 9K^2}}{r} \implies$

$\dfrac{dS}{dr} = \dfrac{2\pi^2 r^6 - 9K^2}{r^2\sqrt{\pi^2 r^6 + 9K^2}} = 0$ $r \neq 0 \implies r = (\dfrac{3K}{\sqrt{2}\pi})^{\frac{1}{3}} \implies h = (\dfrac{6k}{\pi})^{\frac{1}{3}}$

$\implies \tan(\dfrac{\theta}{2}) = \dfrac{h}{r} = (2\sqrt{2})^{\frac{1}{3}} = \sqrt{2} \implies \theta = 2\arctan(\sqrt{2}) \approx 109.47122$ .

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Let $x(t) = at^2 + b,\:\ y(t) = ct^3 + d \implies \dfrac{dy}{dx}|(t = t_{1}) = \dfrac{3c}{2a}t_{1} \implies$ the tangent line to the curve at $(x(t_{1}),y(t_{1}))$ is: $y - (ct_{1}^3 + d) = \dfrac{3c}{2a}t_{1}(x - (at_{1}^2+ b))$

Let the line be normal to the curve at $(x(t_{2}),y(t_{2})) \implies c(t_{2} - t_{1})(t_{2}^2 + t_{1}t_{2} + t_{1}^2) = \dfrac{3c}{2}t_{1}(t_{2} - t_{1})(t_{2} + t_{1}) \implies \dfrac{c}{2}(t_{2} - t_{1})(2t_{2}^2 - t_{1}t_{2} - t_{1}^2) = 0$ $t_{1} \neq t_{2} \implies t_{2} = -\dfrac{t_{1}}{2}$

Since the tangent is also normal to the curve at $(x(t_{2}),y(t_{2})) \implies \dfrac{9c^2}{4a^2}t_{1}t_{2} = -1$ $\implies \dfrac{9c^2}{8a^2}t_{1}^2 = 1 \implies t_{1} = \pm\dfrac{2\sqrt{2}a}{3c} \implies$ the two slopes are $\pm\sqrt{2}$ .

$\tan(\dfrac{\lambda}{2}) = \sqrt{2} \implies \lambda = 2\arctan(\sqrt{2}) \approx \boxed{109.47122}$ .

$\implies \dfrac{\lambda}{\theta} = \boxed{1}$ .

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