Let S be the set of all the natural numbers, for which the line x/a + y/b = 2 is a tangent to the cu

Q: Why does the parameter $n$ disappear in the slope calculation?

That is the beauty of this problem! When you differentiate the curve and substitute the point $(a, b)$, the terms involving $n$ become proportional to each other in such a way that the $n$ factor appears in both the numerator and denominator, allowing it to cancel out. It reveals that the slope at $(a, b)$ is independent of the power $n$.

Q: Can we use the general formula for the tangent to a curve $f(x, y) = c$?

Absolutely! If you are familiar with the gradient method, the slope of the tangent at $(x_0, y_0)$ is given by $-\frac{f_x(x_0, y_0)}{f_y(x_0, y_0)}$. Applying this to our curve will yield the same result, but performing the implicit differentiation manually is a great way to build your intuition for the chain rule.

Q: Is it possible that for some $n$, the line is not a tangent?

Not in this case. Because the slope calculation results in a value that is independent of $n$, and the point $(a, b)$ is always on the curve for any natural number $n$, the line $\frac{x}{a} + \frac{y}{b} = 2$ will always be the tangent line at $(a, b)$ for every $n \in \mathbb{N}$.

Question

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 $\left(\frac{x}{a}ight)^n + \left(\frac{y}{b}ight)^n = 2$ at the point (a, b), $ab 
eq 0$. Then:

Options:

(A) $S = \phi$

(B) $n(S) = 1$

(C) $S = \{2k : k \in \mathbb{N}\}$

(D) $S = \mathbb{N}$

Sigma Percentile · Accepted Answer

### The Geometry of Invariance: A Tangent Journey

Have you ever encountered a problem that looks intimidatingly complex, only to find that it collapses into something beautifully simple? That is exactly what we have here.

We are looking at a family of curves defined by the equation:

$$ \left( \frac{x}{a} ight)^n + \left( \frac{y}{b} ight)^n = 2 $$

We want to determine the conditions under which the line $\frac{x}{a} + \frac{y}{b} = 2$ acts as a tangent at the point $(a, b)$. Let us embark on this journey together.

### Phase 1: The Verification

Before we start applying calculus, we must ensure our foundation is solid. Does the point $(a, b)$ actually lie on our curve?

Let us test it by substituting $x = a$ and $y = b$ into the curve's equation:

$$ \left( \frac{a}{a} ight)^n + \left( \frac{b}{b} ight)^n = 1^n + 1^n = 1 + 1 = 2 $$

It works! The point $(a, b)$ is a permanent resident on this curve, regardless of what natural number $n$ we choose. This is our first clue that the geometry is more stable than the variable $n$ suggests.

### Phase 2: The Calculus of Tangency

To find the tangent, we need the slope. We turn to implicit differentiation of the curve with respect to $x$.

Applying the chain rule, the derivative of the first term is $n \left( \frac{x}{a} ight)^{n-1} \cdot \frac{1}{a}$, and the derivative of the second term is $n \left( \frac{y}{b} ight)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx}$. Since the derivative of the constant $2$ is $0$, we obtain:

$$ n \left( \frac{x}{a} ight)^{n-1} \cdot \frac{1}{a} + n \left( \frac{y}{b} ight)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0 $$

### Phase 3: The Revelation

Now, we evaluate this at our point of interest, $(a, b)$. Substituting $x = a$ and $y = b$, the terms $\left( \frac{a}{a} ight)^{n-1}$ and $\left( \frac{b}{b} ight)^{n-1}$ both become $1^{n-1}$, which is simply $1$.

The equation simplifies dramatically to:

$$ \frac{n}{a} + \frac{n}{b} \frac{dy}{dx} = 0 $$

Look closely—the $n$ is present in both terms. We can divide the entire equation by $n$ (since $n \in \mathbb{N}$, $n 
eq 0$), and it vanishes entirely! We are left with $\frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0$, which yields the slope:

$$ \frac{dy}{dx} = -\frac{b}{a} $$

### Phase 4: The Final Synthesis

With the slope $m = -\frac{b}{a}$ and the point $(a, b)$, we use the point-slope form:

$$ y - b = -\frac{b}{a}(x - a) $$

Multiplying by $a$, we get $a(y - b) = -b(x - a)$, which expands to $ay - ab = -bx + ab$. Rearranging the terms, we find:

$$ bx + ay = 2ab $$

Dividing by $ab$, we arrive at the final form:

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

This is exactly the line given in the problem. Because the $n$ canceled out, this result holds true for any natural number $n$.

Thus, the set $S$ is simply the set of all natural numbers $\mathbb{N}$. You have just proven that this geometric property is **invariant**.

Animated Solution for Mathematics - Differentiation: 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), $ab \neq = 0$ . Then:

Select Answer:

Visualized Solution

Understanding the Problem

Verifying the Point $(a, b)$

Differentiating the Curve

Applying the Chain Rule

Finding the Slope at $(a, b)$

Calculating the Final Slope

Equation of the Tangent Line

Simplifying to Intercept Form

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Invariance

Phase 1

Phase 2

Phase 3

Phase 4

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