Question:

<p><strong>Step 1:</strong> The line is x + y = 2, which has slope = -1. At the point of tangency (N, P), the curve must also have slope -1.</p><p><strong>Step 2:</strong> Find the derivative of the curve $\frac{x^n}{a^n} + \frac{y^n}{b^n} = 2$ using implicit differentiation:</p><p>$\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \cdot \frac{dy}{dx} = 0$</p><p>$\frac{dy}{dx} = -\frac{b^n x^{n-1}}{a^n y^{n-1}}$</p><p><strong>Step 3:</strong> At point (N, P), set the slope equal to -1:</p><p>$-\frac{b^n N^{n-1}}{a^n P^{n-1}} = -1$</p><p>$\frac{b^n N^{n-1}}{a^n P^{n-1}} = 1$</p><p>$b^n N^{n-1} = a^n P^{n-1}$ ... (i)</p><p><strong>Step 4:</strong> The point (N, P) lies on the line x + y = 2:</p><p>$N + P = 2$ ... (ii)</p><p><strong>Step 5:</strong> The point (N, P) lies on the curve:</p><p>$\frac{N^n}{a^n} + \frac{P^n}{b^n} = 2$ ... (iii)</p><p><strong>Step 6:</strong> From equation (i): $\frac{N^{n-1}}{P^{n-1}} = \frac{a^n}{b^n}$, which gives $\frac{N}{P} = \frac{a}{b}$ (taking the n-th root)</p><p><strong>Step 7:</strong> Substituting into equation (ii): $N + P = 2$ and $\frac{N}{P} = \frac{a}{b}$ gives $N = a$ and $P = b$ (when a + b = 2).</p><p><strong>Step 8:</strong> Verify: If N = a and P = b, then $\frac{a^n}{a^n} + \frac{b^n}{b^n} = 1 + 1 = 2$ ✓, and a + b = 2 ✓</p><p><strong>∴ Answer:</strong> B</p>