The curve left(frac{x}{a}right)^n+left(frac{y | vedclass

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

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$\frac{x}{a}+\frac{y}{b}=2$

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(D) Given the curve $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$. To find the slope of the tangent at $(a, b)$,we differentiate with respect to $x$: $n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$. At the point $(a, b)$,we have: $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$. $\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0$. $\frac{dy}{dx} = -\frac{b}{a}$. The equation of the line passing through $(a, b)$ with slope $m = -\frac{b}{a}$ is: $y - b = -\frac{b}{a}(x - a)$. $ay - ab = -bx + ab$. $bx + ay = 2ab$. Dividing by $ab$,we get: $\frac{x}{a} + \frac{y}{b} = 2$.

The curve $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$,where $n \in N$,touches the line at the point $(a, b)$. Find the equation of the line.

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