The equation of the tangent to the curve (fra | vedclass

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(B) Given the curve equation: $(\frac{x}{a})^n+(\frac{y}{b})^n=2$. On differentiating with respect to $x$,we get: $\frac{n x^{n-1}}{a^n} + \frac{n y^{

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

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(B) Given the curve equation: $(\frac{x}{a})^n+(\frac{y}{b})^n=2$. On differentiating with respect to $x$,we get: $\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \frac{dy}{dx} = 0$. $\Rightarrow \frac{dy}{dx} = -\frac{b^n x^{n-1}}{a^n y^{n-1}}$. At the point $(a, b)$,the slope of the tangent is: $(\frac{dy}{dx})_{(a, b)} = -\frac{b^n a^{n-1}}{a^n b^{n-1}} = -\frac{b}{a}$. The equation of the tangent at $(a, b)$ is given by $y - y_1 = m(x - x_1)$: $y - b = -\frac{b}{a}(x - a)$. $ay - ab = -bx + ab$. $bx + ay = 2ab$. Dividing both sides by $ab$,we get: $\frac{x}{a} + \frac{y}{b} = 2$.

The equation of the tangent to the curve $(\frac{x}{a})^n+(\frac{y}{b})^n=2$ at the point $(a, b)$ is

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