Find the value of |a| for which the area of triangle included between the coordinate axes and any tangent to the curve `x ^(a) y = lamda ^(a)` is constant (where `lamda` is constnat.),

Answer

To find the value of |a| for which the area of the triangle included between the coordinate axes and any tangent to the curve \( x^a y = \lambda^a \) is constant, we will follow these steps:

### Step 1: Differentiate the curve
The given curve is:
\[ x^a y = \lambda^a \]
To find the slope of the tangent, we differentiate this equation implicitly with respect to \( x \).

Using the product rule:
\[ \frac{d}{dx}(x^a y) = \frac{d}{dx}(\lambda^a) \]
This gives:
\[ a x^{a-1} y + x^a \frac{dy}{dx} = 0 \]
Rearranging for \(\frac{dy}{dx}\):
\[ x^a \frac{dy}{dx} = -a x^{a-1} y \]
\[ \frac{dy}{dx} = -\frac{a y}{x} \]

### Step 2: Equation of the tangent line
Let the point of tangency be \( (x_1, y_1) \). The slope at this point is:
\[ \frac{dy}{dx} = -\frac{a y_1}{x_1} \]
The equation of the tangent line at this point is:
\[ y - y_1 = -\frac{a y_1}{x_1} (x - x_1) \]

### Step 3: Find the x-intercept
To find the x-intercept, set \( y = 0 \):
\[ 0 - y_1 = -\frac{a y_1}{x_1} (x - x_1) \]
Solving for \( x \):
\[ y_1 = \frac{a y_1}{x_1} (x - x_1) \]
\[ x = x_1 + \frac{x_1}{a} \]
Thus, the x-intercept is:
\[ \left( x_1 \left( 1 + \frac{1}{a} \right), 0 \right) \]

### Step 4: Find the y-intercept
To find the y-intercept, set \( x = 0 \):
\[ y - y_1 = -\frac{a y_1}{x_1} (0 - x_1) \]
This simplifies to:
\[ y = y_1 (1 + a) \]
Thus, the y-intercept is:
\[ \left( 0, y_1 (1 + a) \right) \]

### Step 5: Area of the triangle
The area \( A \) of the triangle formed by the intercepts is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \left( x_1 \left( 1 + \frac{1}{a} \right) \right) \left( y_1 (1 + a) \right) \]
Substituting \( y_1 \) from the curve:
\[ y_1 = \frac{\lambda^a}{x_1^a} \]
Thus, the area becomes:
\[ A = \frac{1}{2} \left( x_1 \left( 1 + \frac{1}{a} \right) \right) \left( \frac{\lambda^a}{x_1^a} (1 + a) \right) \]
Simplifying this gives:
\[ A = \frac{1}{2} \cdot \frac{\lambda^a (1 + a)}{x_1^{a-1} (1 + \frac{1}{a})} \]

### Step 6: Setting the area to be constant
For the area \( A \) to be constant, the term involving \( x_1^{a-1} \) must not depend on \( x_1 \). Therefore, we need:
\[ a - 1 = 0 \implies a = 1 \]

### Conclusion
The value of \( |a| \) for which the area of the triangle is constant is:
\[ |a| = 1 \]