For the curve frac{x^n}{a^n}+frac{y^n}{b^n}=2 | vedclass

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(C) Given the curve $\frac{x^n}{a^n} + \frac{y^n}{b^n} = 2$. At the point $(a, b)$, we check if it lies on the curve: $\frac{a^n}{a^n} + \frac{b^n}{b^

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a tangent for all values of $n$

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(C) Given the curve $\frac{x^n}{a^n} + \frac{y^n}{b^n} = 2$. At the point $(a, b)$, we check if it lies on the curve: $\frac{a^n}{a^n} + \frac{b^n}{b^n} = 1 + 1 = 2$. Thus, $(a, b)$ is on the curve. Differentiating the equation with respect to $x$: $\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \frac{dy}{dx} = 0$. So, $\frac{dy}{dx} = -\frac{x^{n-1}}{a^n} \cdot \frac{b^n}{y^{n-1}} = -\frac{b^n x^{n-1}}{a^n y^{n-1}}$. At the point $(a, b)$, the slope of the tangent $m = -\frac{b^n a^{n-1}}{a^n b^{n-1}} = -\frac{b}{a}$. The equation of the tangent line at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$, which simplifies to $ay - ab = -bx + ab$, or $bx + ay = 2ab$. Dividing by $ab$, we get $\frac{x}{a} + \frac{y}{b} = 2$. Since this matches the given line, the line is a tangent for all values of $n > 1$.

For the curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2, (n \in N \text{ and } n > 1)$, the line $\frac{x}{a}+\frac{y}{b}=2$ is

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