Let S be the set of all natural numbers n for | vedclass

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(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. To find the slope of the tangent at $(a, b)$,we different

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$S = N$

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(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. To find the slope of the tangent at $(a, b)$,we differentiate with respect to $x$: $n \left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} + n \left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \frac{dy}{dx} = 0$. At the point $(a, b)$,we have $\frac{x}{a} = 1$ and $\frac{y}{b} = 1$. Substituting these values: $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \frac{dy}{dx} = 0$. $\frac{n}{a} + \frac{n}{b} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b}{a}$. The equation of the tangent line at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$. $y - b = -\frac{b}{a}x + b \implies \frac{b}{a}x + y = 2b$. Dividing by $b$: $\frac{x}{a} + \frac{y}{b} = 2$. This equation is independent of $n$ and holds for all $n \in N$. Thus,$S = N$.

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

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