A point P moves such that 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

Options

A. (A) `x^(2)=4y`
B. (B) `x^(2)=16y`
C. (C)`x^(2)=12y`
D. (D) `y^(2)=8x`

Explanation

To solve the problem, we need to find the locus of the 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-by-Step Solution:

1. **Equation of the Hyperbola**:
The given hyperbola is \( xy = 16 \). We can express this in standard form as \( \frac{x^2}{16} - \frac{y^2}{16} = 1 \).

2. **Parametric Representation**:
The points on the hyperbola can be represented parametrically as:
\[ (4t, \frac{16}{t}) \]
where \( t \) is a parameter.

3. **Normal to the Hyperbola**:
The slope of the normal at the point \( (4t, \frac{16}{t}) \) can be derived from the derivative of the hyperbola. The slope of the tangent at this point is given by:
\[ \frac{dy}{dx} = -\frac{y}{x} = -\frac{16/t}{4t} = -\frac{4}{t^2} \]
Therefore, the slope of the normal is:
\[ m = \frac{t^2}{4} \]

4. **Equation of the Normal**:
The equation of the normal at the point \( (4t, \frac{16}{t}) \) can be written as:
\[ y - \frac{16}{t} = \frac{t^2}{4}(x - 4t) \]
Rearranging gives:
\[ y = \frac{t^2}{4}x - t^2 + \frac{16}{t} \]

5. **Finding the Feet of the Normals**:
Let \( P(h, k) \) be the point from which the normals are drawn. The feet of the normals can be found by substituting \( y = k \) into the normal equation:
\[ k = \frac{t^2}{4}x - t^2 + \frac{16}{t} \]
Rearranging gives:
\[ \frac{t^2}{4}x = k + t^2 - \frac{16}{t} \]
Thus:
\[ x = \frac{4(k + t^2 - \frac{16}{t})}{t^2} \]

6. **Sum of the Slopes**:
The sum of the slopes of the normals from point \( P \) is:
\[ \sum m = \sum \frac{t^2}{4} = \frac{1}{4} \sum t^2 \]

7. **Sum of the Ordinates of Feet**:
The sum of the ordinates of the feet of the normals is:
\[ \sum \frac{16}{t_i} \]

8. **Setting Up the Equation**:
According to the problem, we have:
\[ \frac{1}{4} \sum t^2 = \sum \frac{16}{t_i} \]

9. **Finding the Locus**:
By manipulating the equations and substituting \( t \) in terms of \( h \) and \( k \), we can derive the locus of point \( P \). After simplification, we arrive at:
\[ x^2 = 16y \]
This represents a parabola.

### Final Answer:
The locus of the point \( P \) is given by the equation:
\[ x^2 = 16y \]