Prove that $\left(\frac{\mathrm{x}}{\mathrm{a}}\right)^{\mathrm{n}}+\left(\frac{\mathrm{y}}{\mathrm{b}}\right)^{\mathrm{n}}=2$ touches the straight line $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=2$ for all $\mathrm{n} \in \mathrm{N}$, at the point $(\mathrm{a}, \mathrm{b})$.
finding the slope of the tangent by differentiating the curve
$n\left(\frac{x}{a}\right)^{n-1}+n\left(\frac{y}{b}\right)^{n-1} \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\left(\frac{x}{y}\right)^{n-1}\left(\frac{b}{a}\right)^{n}$
$\mathrm{m}$ (tangent) at $(\mathrm{a}, \mathrm{b})$ is $-\frac{\mathrm{b}}{\mathrm{a}}$
equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$
therefore, the equation of the tangent is
$y-b=-\frac{b}{a}(x-a)$
$\frac{x}{a}+\frac{y}{b}=2$
Hence, proved
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