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