Common Tangents

$f'(a) = (2n + 1)a^{2n} = g'(b) = \dfrac{1}{(2n + 1)b^{\frac{2n}{2n + 1}}} \implies$ $a = \pm\dfrac{1}{\alpha(n)b^{\frac{1}{2n + 1}}}$ , where $\alpha(n) = (2n + 1)^{\frac{1}{n}}$

Using $a = -\dfrac{1}{\alpha(n)b^{\frac{1}{2n + 1}}} \implies$

$A:(-\dfrac{1}{\alpha(n)b^{\frac{1}{2n + 1}}}, -\dfrac{1}{\alpha(n)^{2n + 1}b})$ and $B:(b,b^{\frac{1}{2n + 1}})$

$\implies m_{AB} = (\dfrac{\alpha(n)^{2n + 1}b^{\frac{2n + 2}{2n + 1}} + 1}{\alpha(n)b^{\frac{2n + 2}{2n + 1}} + 1})(\dfrac{1}{\alpha(n)^{2n}b^{\frac{2n}{2n + 1}}}) = \dfrac{1}{\alpha(n)b^{\frac{2n}{2n + 1}}} \implies \alpha(n)^{n + 1}b^{\frac{2(n + 1)}{2n + 1}} = 1 \implies b = (\dfrac{1}{\alpha(n)})^{\frac{2n + 1}{2}} \implies$ $a = (\dfrac{1}{\alpha(n)})^{\frac{1}{2}}$

$\alpha(n) = (2n + 1)^{\frac{1}{n}} \implies b = \dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}}$ and $a = \dfrac{1}{(2n + 1)^{\frac{1}{2n}}}$

$\implies m_{AB} = g'(b) = 1$ and $A:(-\dfrac{1}{(2n + 1)^{\frac{1}{2n}}},-\dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}})$

Using the symmetry about the line $y = x \implies A'(-\dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}},-\dfrac{1}{(2n + 1)^{\frac{1}{2n}}})$

$\implies$ the distance $d = AA' = \dfrac{2n\sqrt{2}}{(2n + 1)^{\frac{2n + 1}{2n}}} = \dfrac{\lambda(n)\sqrt{2}}{\gamma(n)^{\frac{\gamma(n)}{\lambda(n)}}}$

and,

$\gamma(n) + \lambda(n) = (2n + 1) + (2n) = 4n + 1 = 9 \implies n = \boxed{2}$ .

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