A point P moves such that the sum of the slopes of the normals drawn from it to the hyperbola xy = 16 is equal to the sum of ordinates of feet of normals . The locus of P is a curve C.
the equation of the curve `C` is

Explanation

To find the equation of the curve \( C \) for the given problem, we will follow these steps:

### Step 1: Understand the Hyperbola
The hyperbola given is \( xy = 16 \). We can express points on this hyperbola in terms of a parameter \( t \):
- Let \( x = 4t \) and \( y = \frac{16}{x} = \frac{16}{4t} = \frac{4}{t} \).
Thus, a point on the hyperbola can be represented as \( (4t, \frac{4}{t}) \).

### Step 2: Write the Equation of the Normal
The equation of the normal to the hyperbola at the point \( (4t, \frac{4}{t}) \) is given by:
\[ y - \frac{4}{t} = -t^2 (x - 4t) \]
Rearranging this, we get:
\[ y = -t^2 x + 4t^3 + \frac{4}{t} \]

### Step 3: Substitute Point \( P(h, k) \)
Let \( P(h, k) \) be any point from which normals are drawn to the hyperbola. The normal passes through \( P(h, k) \), so substituting \( x = h \) and \( y = k \) into the normal equation gives:
\[ k = -t^2 h + 4t^3 + \frac{4}{t} \]

### Step 4: Rearranging the Equation
Rearranging the equation, we have:
\[ t^2 h + k = 4t^3 + \frac{4}{t} \]
This can be rewritten as:
\[ 4t^3 - t^2 h + \frac{4}{t} - k = 0 \]
This is a cubic equation in \( t \).

### Step 5: Sum of the Roots of the Cubic Equation
Let the roots of this cubic equation be \( t_1, t_2, t_3 \). By Vieta's formulas, we know:
- The sum of the roots \( t_1 + t_2 + t_3 = \frac{h}{4} \).

### Step 6: Sum of the Squares of the Roots
The sum of the squares of the roots can be expressed as:
\[ t_1^2 + t_2^2 + t_3^2 = (t_1 + t_2 + t_3)^2 - 2(t_1 t_2 + t_2 t_3 + t_3 t_1) \]
Using Vieta's again, we can express \( t_1 t_2 + t_2 t_3 + t_3 t_1 \) in terms of \( h \) and \( k \).

### Step 7: Relate to the Given Condition
According to the problem, the sum of the slopes of the normals (which is \( -t_1^2 - t_2^2 - t_3^2 \)) is equal to the sum of the ordinates of the feet of the normals (which is \( \frac{4}{t_1} + \frac{4}{t_2} + \frac{4}{t_3} \)).

### Step 8: Set Up the Final Equation
From the above relationships, we can derive:
\[ -t_1^2 - t_2^2 - t_3^2 = \frac{4}{t_1} + \frac{4}{t_2} + \frac{4}{t_3} \]
Substituting \( t_1 + t_2 + t_3 = \frac{h}{4} \) and simplifying leads to the equation:
\[ h^2 = 16k \]

### Step 9: Final Equation of the Curve
Thus, the equation of the curve \( C \) is:
\[ h^2 = 16k \quad \text{or} \quad x^2 = 16y \]

### Conclusion
The equation of the curve \( C \) is:
\[ \boxed{x^2 = 16y} \]