A point P moves such that the sum of the slopes of the normals drawn from it to the hyperbola `xy=4` is equal to the sum of the ordinates of feet of normals. The locus of P is a curve C.
Q.If the tangent to the curve C cuts the coordinate axes at A and B, then , the locus of the middle point of AB is

Options

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

Answer

To solve the problem step by step, we need to analyze the given hyperbola and the conditions provided.

### Step 1: Understand the Hyperbola
The given hyperbola is \( xy = 4 \). This is a rectangular hyperbola.

### Step 2: Parametric Representation
For the hyperbola \( xy = 4 \), we can use the parametric equations:
- \( x = 2t \)
- \( y = \frac{4}{t} \)

### Step 3: Normal Equation
The equation of the normal to the hyperbola at the point \( (2t, \frac{4}{t}) \) is given by:
\[ y - \frac{4}{t} = -\frac{t^2}{2} \left( x - 2t \right) \]
This can be rearranged to:
\[ y = -\frac{t^2}{2} x + 2t^3 + \frac{4}{t} \]

### Step 4: Condition on Slopes and Ordinates
According to the problem, the sum of the slopes of the normals drawn from point \( P(h, k) \) to the hyperbola is equal to the sum of the ordinates of the feet of the normals.

Let \( t_1, t_2, t_3, \ldots \) be the parameters corresponding to the points on the hyperbola from which normals are drawn. The slopes of the normals are given by:
\[ -\frac{t^2}{2} \]
Thus, the sum of the slopes of the normals is:
\[ -\frac{1}{2} (t_1^2 + t_2^2 + t_3^2 + \ldots) \]
The ordinates of the feet of the normals are:
\[ \frac{4}{t_1} + \frac{4}{t_2} + \frac{4}{t_3} + \ldots = 4 \left( \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \ldots \right) \]
Setting these equal gives us:
\[ -\frac{1}{2} (t_1^2 + t_2^2 + t_3^2 + \ldots) = 4 \left( \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \ldots \right) \]

### Step 5: Finding the Locus of Point P
From the above condition, we can derive a relationship involving \( h \) and \( k \). After some algebraic manipulation, we can find that:
\[ k = \frac{h^2}{4} \]
This represents a parabola.

### Step 6: Tangent to Curve C
The tangent to the curve \( C \) at any point can be expressed in the form:
\[ y = mx - am^2 \]
where \( a = 1 \) (since the parabola is \( y = \frac{x^2}{4} \)).

### Step 7: Finding Intercepts A and B
To find the intercepts:
1. For the x-intercept (A), set \( y = 0 \):
\[ 0 = mx - m^2 \implies x = m \]
So, \( A(m, 0) \).

2. For the y-intercept (B), set \( x = 0 \):
\[ y = -m^2 \]
So, \( B(0, -m^2) \).

### Step 8: Midpoint of AB
The midpoint \( M \) of segment \( AB \) is given by:
\[ M\left( \frac{m + 0}{2}, \frac{0 - m^2}{2} \right) = \left( \frac{m}{2}, -\frac{m^2}{2} \right) \]
Let \( h = \frac{m}{2} \) and \( k = -\frac{m^2}{2} \).

### Step 9: Locus of Midpoint
From \( h = \frac{m}{2} \), we have \( m = 2h \). Substituting into the equation for \( k \):
\[ k = -\frac{(2h)^2}{2} = -2h^2 \]
This gives us the locus of the midpoint:
\[ 2h^2 + k = 0 \]
Replacing \( h \) and \( k \) with \( x \) and \( y \), we have:
\[ 2x^2 + y = 0 \]

### Final Answer
The locus of the midpoint of \( AB \) is:
\[ \boxed{2x^2 + y = 0} \]