Let the function `f : [-4,4] to [-1, 1]` be defined implicitly by the equation `x +5y - y^(5)=0` If the area of triangle formed by tangent and normal to `f (x) at x=0` and the line `y =5` is A, find `(A)/(13).`

Answer

To solve the problem, we need to find the area of the triangle formed by the tangent and normal to the function \( f(x) \) at \( x = 0 \) and the line \( y = 5 \). We will follow these steps:

### Step 1: Differentiate the Implicit Function
We start with the implicit function given by the equation:
\[ x + 5y - y^5 = 0 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x) + 5\frac{dy}{dx} - 5y^4\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 1 + (5 - 5y^4)\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{1}{5 - 5y^4} \]

### Step 2: Find the Value of \( y \) at \( x = 0 \)
Substituting \( x = 0 \) into the original equation:
\[ 0 + 5y - y^5 = 0 \implies 5y = y^5 \implies y(y^4 - 5) = 0 \]
Thus, \( y = 0 \) or \( y = 5^{1/4} \). Since \( 5^{1/4} \) is not in the range \([-1, 1]\), we take \( y = 0 \).

### Step 3: Calculate the Slope of the Tangent Line at \( x = 0 \)
Substituting \( y = 0 \) into the derivative:
\[ \frac{dy}{dx} = \frac{1}{5 - 5(0)^4} = \frac{1}{5} \]
So, the slope of the tangent line at \( (0, 0) \) is \( \frac{1}{5} \).

### Step 4: Write the Equation of the Tangent Line
Using the point-slope form of the line:
\[ y - 0 = \frac{1}{5}(x - 0) \implies y = \frac{1}{5}x \]

### Step 5: Calculate the Slope of the Normal Line
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -5 \]
Thus, the equation of the normal line is:
\[ y - 0 = -5(x - 0) \implies y = -5x \]

### Step 6: Find Points of Intersection with the Line \( y = 5 \)
1. For the tangent line \( y = \frac{1}{5}x \):
\[ 5 = \frac{1}{5}x \implies x = 25 \implies (25, 5) \]
2. For the normal line \( y = -5x \):
\[ 5 = -5x \implies x = -1 \implies (-1, 5) \]

### Step 7: Area of the Triangle Formed
The vertices of the triangle are \( (25, 5) \), \( (-1, 5) \), and \( (0, 0) \). The base of the triangle is the distance between \( (25, 5) \) and \( (-1, 5) \):
\[ \text{Base} = 25 - (-1) = 26 \]
The height of the triangle is the vertical distance from the point \( (0, 0) \) to the line \( y = 5 \), which is \( 5 \).

The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 26 \times 5 = 65 \]

### Step 8: Find \( \frac{A}{13} \)
Now, we need to find:
\[ \frac{A}{13} = \frac{65}{13} = 5 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{5} \]