The curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2$ touches the line $\frac{x}{a}+\frac{y}{b}=2$ at the point 

$(a, b)$

We have,

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

Differentiating with respect to x, we get

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

Suppose the line $\frac{x}{a}+\frac{y}{b}=1$ touches the curve (i) at $P(\alpha, \beta)$.

The equation of the tangent at $P(\alpha, \beta)$ is

$y-\beta=-\frac{b^n}{a^n}\left(\frac{\alpha}{\beta}\right)^{n-1} (x-\alpha)$

$\Rightarrow a^n \beta^{n-1} y-a^n \beta^n=-b^n \alpha^{n-1} x+b^n \alpha^n$

$\Rightarrow \left(b^n \alpha^{n-1}\right) x+\left(a^n \beta^{n-1}\right) y=a^n \beta^n+b^n \alpha^n$

$\Rightarrow \left(b^n \alpha^{n-1}\right) x+\left(a^n \beta^{n-1}\right) y=2 a^n b^n$      [∵  (α, β) lies on (i)]

Comparing this equation with $\frac{x}{a}+\frac{y}{b}=2$, we get

$\frac{a b^n \alpha^{n-1}}{a^n b^n}=\frac{b a^n \beta^{n-1}}{a^n b^n}=\frac{2}{2}$

$\Rightarrow \left(\frac{\alpha}{a}\right)^{n-1}=\left(\frac{\beta}{b}\right)^{n-1}=1$

$\Rightarrow \frac{\alpha}{a}=\frac{\beta}{b}=1 \Rightarrow \alpha=a, \beta=b$