Let S be the set of all the natural numbers, for which the line ${x \over a} + {y \over b} = 2$ is a tangent to the curve ${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$ at the point (a, b), ab $\ne$ 0. Then :

<p>${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$</p> <p>Differentiating both sides with respect to x, we get</p> <p>$$ \Rightarrow n\,.\,{\left( {{x \over a}} \right)^{n - a}}\,.\,{1 \over a} + n\,.\,{\left( {{y \over b}} \right)^{n - 1}}\,.\,{1 \over b}\,.\,{{dy} \over {dx}} = 0$$</p> <p>$$ \Rightarrow {{dy} \over {dx}} = - {{{1 \over a}\,.\,{{\left( {{x \over a}} \right)}^{n - 1}}} \over {{1 \over b}{{\left( {{y \over b}} \right)}^{n - 1}}}}$$</p> <p>$= - {b \over a}{\left( {{{xb} \over {ya}}} \right)^{n - 1}}$</p> <p>Now, ${{dy} \over {dx}}$ at (a, b) $= - {b \over a}{\left[ {{{ab} \over {ba}}} \right]^{n - 1}} = - {b \over a}$</p> <p>Equation of tangent at (a, b) is,</p> <p>$(y - b) = - {b \over a}(x - a)$</p> <p>$\Rightarrow {{y - b} \over b} = - {{x - a} \over a}$</p> <p>$\Rightarrow {y \over b} - 1 = - {x \over a} + 1$</p> <p>$\Rightarrow {x \over a} + {y \over b} = 2$</p> <p>$\therefore$ It is tangent for all value of n.</p>