Find the value of `n in N` such that the curve `(x/a)^n+(y/b)^n=2` touches the straight line `x/a+y/b=2` at the point `(a , b)dot`

Answer

To solve the problem, we need to find the value of \( n \) in the natural numbers such that the curve given by

\[ \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \]

touches the straight line

\[ \frac{x}{a} + \frac{y}{b} = 2 \]

at the point \( (a, b) \).

### Step 1: Differentiate the curve equation

We start by differentiating the curve equation with respect to \( x \):

\[ \frac{d}{dx}\left(\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n\right) = 0 \]

Using the chain rule, we get:

\[ n \left(\frac{1}{a}\right) \left(\frac{x}{a}\right)^{n-1} + n \left(\frac{1}{b}\right) \left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)

Rearranging the equation gives:

\[ n \left(\frac{1}{b}\right) \left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = -n \left(\frac{1}{a}\right) \left(\frac{x}{a}\right)^{n-1} \]

Dividing both sides by \( n \) (assuming \( n \neq 0 \)):

\[ \left(\frac{1}{b}\right) \left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = -\left(\frac{1}{a}\right) \left(\frac{x}{a}\right)^{n-1} \]

Thus, we can express \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{\left(\frac{x}{a}\right)^{n-1}}{\left(\frac{y}{b}\right)^{n-1}} \]

### Step 3: Evaluate the derivative at the point \( (a, b) \)

Now we substitute \( x = a \) and \( y = b \):

\[ \frac{dy}{dx} \bigg|_{(a,b)} = -\frac{b}{a} \cdot \frac{\left(\frac{a}{a}\right)^{n-1}}{\left(\frac{b}{b}\right)^{n-1}} = -\frac{b}{a} \]

### Step 4: Find the equation of the tangent line at \( (a, b) \)

The equation of the tangent line at the point \( (a, b) \) can be written as:

\[ y - b = \left(-\frac{b}{a}\right)(x - a) \]

Rearranging gives:

\[ y - b = -\frac{b}{a}x + \frac{b}{a}a \]

\[ y = -\frac{b}{a}x + 2b \]

### Step 5: Compare with the given line equation

The given line equation is:

\[ \frac{x}{a} + \frac{y}{b} = 2 \]

Rearranging gives:

\[ y = -\frac{b}{a}x + 2b \]

### Conclusion

Since both equations of the tangent line and the given line are identical, we conclude that the line touches the curve at the point \( (a, b) \) for all \( n \) in the natural numbers \( \mathbb{N} \).

Thus, the value of \( n \) can be any natural number.