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

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Question

(A) Given the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$. At the point where the abscissa $x = a$,we have $(\frac{a}{a})^n + (\frac{y}{b})^n = 2$,w

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

Vedclass · Answer

(A) Given the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$. At the point where the abscissa $x = a$,we have $(\frac{a}{a})^n + (\frac{y}{b})^n = 2$,which implies $1 + (\frac{y}{b})^n = 2$,so $(\frac{y}{b})^n = 1$. Since $n \in N$,this gives $y = b$ (assuming $n$ is odd or considering the principal value). Thus,the point of tangency is $(a, b)$. Differentiating the curve with respect to $x$: $n(\frac{x}{a})^{n-1} \cdot \frac{1}{a} + n(\frac{y}{b})^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$. At $(a, b)$,this becomes $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$. $\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b}{a}$. The equation of the tangent at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$. $a(y - b) = -b(x - a) \implies ay - ab = -bx + ab$. $bx + ay = 2ab \implies \frac{x}{a} + \frac{y}{b} = 2$.

The equation of the tangent to the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$ $(n \in N)$ at the point with abscissa equal to '$a$' is:

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