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Answer: To Prove $\frac{x}{a}+\frac{y}{b}=2$(R.H.S=L.H.S)

Hint: Differentiate

Given:$\left [ \frac{x}{a} \right ]^{n}+\left [ \frac{y}{b} \right ]^{n}=2$

Solution:$\left [ \frac{x}{a} \right ]^{n}+\left [ \frac{y}{b} \right ]^{n}=2$

\begin{aligned} &\Rightarrow \frac{n}{a}\left[\frac{x}{a}\right]^{n-1}+\frac{n}{b}\left[\frac{y}{b}\right]^{n-1} \frac{d y}{d x}=0 \\ &\Rightarrow \frac{n}{b}\left[\frac{y}{b}\right]^{n-1} \frac{d y}{d x}=-\frac{n}{a}\left[\frac{x}{a}\right]^{n-1} \\ &\Rightarrow \frac{d y}{d x}=-\frac{n}{a}\left[\frac{x}{a}\right]^{n-1} \times \frac{b}{n}\left[\frac{b}{y}\right]^{n-1}=-\frac{b}{a}\left(\frac{b x}{a y}\right)^{n-1} \end{aligned}

$\therefore slope \: of \, tangent=$$\left(\frac{d y}{d x}\right)_{(a, b)}=-\frac{b}{a}\left(\frac{b \times a}{a \times b}\right)^{n-1}=-\frac{b}{a}$....(ii)

The equation of the tangent is

\begin{aligned} &y-b=-\frac{b}{a}(x-a) \\ &\Rightarrow y a-a b=-x b+a b \\ &\Rightarrow x b+y a=2 a b \\ &\Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{aligned}

So the given line touched the given curve at the given point

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