Let S be the set of all the natural numbers, for which the line $\frac{x}{a} + \frac{y}{b} = 2$ is a tangent to the curve ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$ at the point $(a, b), a b \neq 0$ . Then:

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$n (S) = 1$

$S = ϕ$

S=N

$S = {2 k : k \in N}$

answer is D.

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Detailed Solution

${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2 Slope of tangent at (a, b) \begin{aligned} n \cdot {(\frac{x}{a})}^{n - 1} \cdot \frac{1}{a} + n {(\frac{x}{b})}^{n - 1} \cdot \frac{1}{b} \frac{dy}{dx} = 0 \\ {\frac{dy}{dx}|}_{(a, b)} = - \frac{b}{a} \end{aligned} ∴ Equation of tangent is \begin{aligned} y - b = - \frac{b}{a} (x - a) \\ \frac{x}{a} + \frac{y}{b} = 2 \forall n \in N \end{aligned}$