Find the condition that the line `A x+B y=1` may be normal to the curve `a^(n-1)y=x^ndot`

Answer

To find the condition that the line \( Ax + By = 1 \) may be normal to the curve \( a^{n-1}y = x^n \), we will follow these steps:

### Step 1: Identify the point of tangency
Let the point of tangency on the curve be \( P(x_1, y_1) \). Therefore, at this point, we have:
\[ a^{n-1}y_1 = x_1^n \tag{1} \]
And since the line is normal to the curve at this point, we also have:
\[ Ax_1 + By_1 = 1 \tag{2} \]

### Step 2: Differentiate the curve
To find the slope of the tangent to the curve at point \( P \), we differentiate the equation of the curve with respect to \( x \):
\[ \frac{d}{dx}(a^{n-1}y) = \frac{d}{dx}(x^n) \]
Using the product rule on the left side:
\[ a^{n-1} \frac{dy}{dx} = nx^{n-1} \]
Thus, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{nx^{n-1}}{a^{n-1}} \tag{3} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{a^{n-1}}{nx^{n-1}} \tag{4} \]

### Step 4: Find the slope of the given line
The slope of the line \( Ax + By = 1 \) can be rearranged to:
\[ y = -\frac{A}{B}x + \frac{1}{B} \]
Thus, the slope of the line is:
\[ \text{slope of line} = -\frac{A}{B} \tag{5} \]

### Step 5: Set slopes equal
Since the line is normal to the curve, we equate the slopes from equations (4) and (5):
\[ -\frac{a^{n-1}}{nx_1^{n-1}} = -\frac{A}{B} \]
This simplifies to:
\[ \frac{a^{n-1}}{nx_1^{n-1}} = \frac{A}{B} \tag{6} \]

### Step 6: Substitute \( y_1 \) from equation (1)
From equation (1), we can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = \frac{x_1^n}{a^{n-1}} \tag{7} \]
Substituting \( y_1 \) into equation (2):
\[ Ax_1 + B\left(\frac{x_1^n}{a^{n-1}}\right) = 1 \]
This gives us:
\[ Ax_1 + \frac{Bx_1^n}{a^{n-1}} = 1 \tag{8} \]

### Step 7: Solve for \( x_1 \)
Rearranging equation (8):
\[ Ax_1 + \frac{Bx_1^n}{a^{n-1}} = 1 \]
Multiply through by \( a^{n-1} \):
\[ a^{n-1}Ax_1 + Bx_1^n = a^{n-1} \]
This can be rewritten as:
\[ Bx_1^n + a^{n-1}Ax_1 - a^{n-1} = 0 \tag{9} \]

### Step 8: Condition for normality
The condition for the line to be normal to the curve is that the discriminant of this polynomial must be non-negative. Therefore, we can find the required condition by analyzing the roots of the polynomial formed in equation (9).

### Final Condition
The condition that the line \( Ax + By = 1 \) is normal to the curve \( a^{n-1}y = x^n \) can be derived from the relationships established in the previous steps.