The equation of normal to the curve `(x/a)^n+(y/b)^n=2(n in N)` at the point with abscissa equal to 'a can be

Options

A. `xa+yb=a^2 - b^2`
B. `xa+yb=a^2 + b^2`
C. `xa - yb=a^2 - b^2`
D. `bx-ay=a^2 - b^2`

Answer

To find the equation of the normal to the curve \((x/a)^n + (y/b)^n = 2\) at the point where the abscissa is equal to \(a\), we will follow these steps:

### Step 1: Find the coordinates of the point on the curve
We substitute \(x = a\) into the curve equation:
\[ \left(\frac{a}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \]
This simplifies to:
\[ 1 + \left(\frac{y}{b}\right)^n = 2 \]
From this, we can solve for \(y\):
\[ \left(\frac{y}{b}\right)^n = 1 \implies y^n = b^n \]
Thus, \(y = b\) or \(y = -b\) (since \(n\) is a natural number, both positive and negative values are valid when \(n\) is even).

### Step 2: Differentiate the curve to find the slope of the tangent
We differentiate the curve implicitly:
\[ \frac{d}{dx}\left(\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n\right) = 0 \]
Using the chain rule, we get:
\[ n\left(\frac{x}{a}\right)^{n-1}\cdot\frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1}\cdot\frac{1}{b}\cdot\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{\left(\frac{x}{a}\right)^{n-1}}{\left(\frac{y}{b}\right)^{n-1}} \]

### Step 3: Substitute the point to find the slope at \(x = a\)
Substituting \(x = a\) and \(y = b\) into the derivative:
\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{(1)^{n-1}}{(1)^{n-1}} = -\frac{b}{a} \]
Thus, the slope of the tangent at the point \((a, b)\) is \(-\frac{b}{a}\).

### Step 4: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = \frac{a}{b} \]

### Step 5: Write the equation of the normal line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (a, b)\) and \(m = \frac{a}{b}\):
\[ y - b = \frac{a}{b}(x - a) \]
Multiplying through by \(b\) to eliminate the fraction:
\[ b(y - b) = a(x - a) \]
This simplifies to:
\[ by - b^2 = ax - a^2 \]
Rearranging gives:
\[ ax - by = a^2 - b^2 \]

### Step 6: Write the second normal equation
For the case where \(y = -b\):
\[ y + b = -\frac{a}{b}(x - a) \]
Multiplying through by \(b\):
\[ b(y + b) = -a(x - a) \]
This simplifies to:
\[ by + b^2 = -ax + a^2 \]
Rearranging gives:
\[ ax + by = a^2 + b^2 \]

### Final Answer
The equations of the normals to the curve at the points \((a, b)\) and \((a, -b)\) are:
1. \(ax - by = a^2 - b^2\)
2. \(ax + by = a^2 + b^2\)