If ab neq 0,then the equation of the tangent | vedclass

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

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

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(C) Given the curve equation is $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$. To find the slope of the tangent at $(a, b)$,we differentiate both sides 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 substitute $x = a$ and $y = b$: $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$. $\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0$. $\frac{n}{b} \cdot \frac{dy}{dx} = -\frac{n}{a} \implies \frac{dy}{dx} = -\frac{b}{a}$. The equation of the tangent line at $(a, b)$ with slope $m = -\frac{b}{a}$ is given by $y - b = m(x - a)$. $y - b = -\frac{b}{a}(x - a)$. $a(y - b) = -b(x - a)$. $ay - ab = -bx + ab$. $bx + ay = 2ab$. Dividing both sides by $ab$,we get $\frac{bx}{ab} + \frac{ay}{ab} = \frac{2ab}{ab}$. $\frac{x}{a} + \frac{y}{b} = 2$. Thus,the correct option is $C$.

If $ab \neq 0$,then the equation of the tangent at $(a, b)$ to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ is

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