The sum of the intercepts made by a tangent d | vedclass

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(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. Taking the derivative with respect to $x$: $\frac{n}{a} \

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$2(a+b)$

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(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. Taking the derivative with respect to $x$: $\frac{n}{a} \left(\frac{x}{a}\right)^{n-1} + \frac{n}{b} \left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = 0$. At the point $(a, b)$,the slope $m$ is: $m = \left(\frac{dy}{dx}\right)_{(a,b)} = -\frac{b}{a} \left(\frac{a/a}{b/b}\right)^{n-1} = -\frac{b}{a}$. The equation of the tangent at $(a, b)$ is: $y - b = -\frac{b}{a}(x - a) \Rightarrow ay - ab = -bx + ab \Rightarrow bx + ay = 2ab$. Dividing by $2ab$,we get: $\frac{x}{2a} + \frac{y}{2b} = 1$. The intercepts on the axes are $2a$ and $2b$. The sum of the intercepts is $2a + 2b = 2(a+b)$.

The sum of the intercepts made by a tangent drawn to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ at the point $(a, b)$ on the coordinate axes is:

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