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

Differentiating, we get, 

$n{\left(\frac{x}{a}\right)}^{n-1}\cdot \frac{1}{a}+n\cdot {\left(\frac{y}{b}\right)}^{n-1}\cdot \frac{1}{b}\cdot \frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-\frac{1}{a}{\left(\frac{x}{a}\right)}^{n-1}}{\frac{1}{b}{\left(\frac{y}{b}\right)}^{n-1}}=\frac{-b}{a}{\left(\frac{xb}{ya}\right)}^{n-1}\Rightarrow {\left.\frac{dy}{dx}\right|}_{\left(a,b\right)}=\frac{-b}{a}{\left(\frac{ab}{ba}\right)}^{n-1}$ $=\frac{-b}{a}$

So, equation of the tangent to the curve at $\left(a,b\right)$ will be $y-b=\frac{-b}{a}\left(x-a\right)$  $\Rightarrow \frac{y}{b}-1=-\frac{x}{a}+1$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

i.e. $S\in N$