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.
If the tangent to the curve C cuts the corrdinate axes at A and B, then the locus of the middle point of AB is

Options

A. `x^(2)=16y`
B. `x^(2)=2y`
C. `x^(2)+2y=0`
D. `x^(2)+4y=0`

Explanation

To solve the problem, we need to find the locus of point \( P(h, k) \) such that the sum of the slopes of the normals drawn from \( P \) to the hyperbola \( xy = 16 \) is equal to the sum of the ordinates of the feet of the normals.

### Step 1: Understanding the Hyperbola
The given hyperbola is \( xy = 16 \). We can express points on the hyperbola in terms of a parameter \( t \):
\[ x = 4t, \quad y = \frac{16}{x} = \frac{16}{4t} = \frac{4}{t} \]
So, the points on the hyperbola can be represented as \( (4t, \frac{4}{t}) \).

### Step 2: Finding the Normal Line
The slope of the tangent to the hyperbola at point \( (4t, \frac{4}{t}) \) can be found using implicit differentiation:
\[ \frac{dy}{dx} = -\frac{y}{x} = -\frac{4/t}{4t} = -\frac{1}{t^2} \]
Thus, the slope of the normal is the negative reciprocal:
\[ \text{slope of normal} = t^2 \]
The equation of the normal line at point \( (4t, \frac{4}{t}) \) is:
\[ k - \frac{4}{t} = t^2(h - 4t) \]

### Step 3: Rearranging the Normal Equation
Rearranging the normal equation gives us:
\[ k - \frac{4}{t} = t^2h - 4t^3 \]
This can be rewritten as:
\[ t^2h - k + 4t^3 - \frac{4}{t} = 0 \]

### Step 4: Sum of Slopes of Normals
Let \( t_1, t_2, t_3, t_4 \) be the slopes of the normals from point \( P \) to the hyperbola. The sum of the slopes is:
\[ t_1 + t_2 + t_3 + t_4 \]

### Step 5: Sum of the Ordinates of Feet of Normals
The ordinates of the feet of the normals can be expressed as:
\[ \frac{4}{t_1} + \frac{4}{t_2} + \frac{4}{t_3} + \frac{4}{t_4} = 4\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4}\right) \]

### Step 6: Setting Up the Equation
According to the problem, we have:
\[ t_1 + t_2 + t_3 + t_4 = 4\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4}\right) \]

### Step 7: Using Symmetric Sums
Using the properties of symmetric sums, we can express the above equation in terms of \( h \) and \( k \):
\[ \text{Let } S_1 = t_1 + t_2 + t_3 + t_4, \quad S_2 = t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4 \]
From Vieta's formulas, we can relate these sums to \( h \) and \( k \).

### Step 8: Finding the Locus
After some algebraic manipulation, we find that:
\[ h^2 = 16k \]
This represents a parabola.

### Step 9: Finding the Locus of Midpoint of AB
If the tangent to curve \( C \) cuts the axes at points \( A \) and \( B \), the midpoint \( M \) of segment \( AB \) can be expressed as:
\[ M\left(\frac{h}{2}, \frac{k}{2}\right) \]
Substituting \( h^2 = 16k \) into this gives us:
\[ \left(\frac{h}{2}\right)^2 = 4k \implies x^2 = 4y \]
This is the equation of a parabola.

### Final Answer
The locus of the midpoint of \( AB \) is:
\[ \boxed{x^2 = 4y} \]