Let S be the set of all the natural numbers, for which the line...

{\left(\dfrac{x}{a}\right)}^{n} + {\left(\dfrac{y}{b}\right)}^{n} = 2$</p><p>Differentiating, we get, </p><p>$n{\left(\dfrac{x}{a}\right)}^{n - 1} ⋅ \dfrac{1}{a} + n ⋅ {\left(\dfrac{y}{b}\right)}^{n - 1} ⋅ \dfrac{1}{b} ⋅ \dfrac{dy}{dx} = 0$</p><p>$\dfrac{dy}{dx} = \dfrac{- \dfrac{1}{a}{\left(\dfrac{x}{a}\right)}^{n - 1}}{\dfrac{1}{b}{\left(\dfrac{y}{b}\right)}^{n - 1}} = \dfrac{- b}{a}{\left(\dfrac{xb}{ya}\right)}^{n - 1} ⇒ {\left\dfrac{dy}{dx}\right|}_{\left(a, b\right)} = \dfrac{- b}{a}{\left(\dfrac{ab}{ba}\right)}^{n - 1} = \dfrac{- b}{a}$</p><p>So, equation of the tangent to the curve at \left(a, b\right)$ will be $y - b = \dfrac{- b}{a}\left(x - a\right) ⇒ \dfrac{y}{b} - 1 = - \dfrac{x}{a} + 1$

$⇒ \dfrac{x}{a} + \dfrac{y}{b} = 2$

i.e. $S \in N$