The locus of mid - point of the portion of a line of constant slope 'm' between two branches of a rectangular hyperbola `xy = 1 ` is

Options

A. `y - mx = 0 `
B. `y + mx = 0 `
C. `my + x = 0 `
D. `y = x `

Answer

To find the locus of the midpoint of a line segment of constant slope \( m \) that lies between the two branches of the rectangular hyperbola defined by \( xy = 1 \), we can follow these steps:

### Step 1: Understand the Hyperbola
The equation of the hyperbola is given by:
\[ xy = 1 \]
This hyperbola has two branches, one in the first quadrant and one in the third quadrant.

### Step 2: Equation of the Line
Let the midpoint of the line segment be \( P(h, k) \). Since the line has a constant slope \( m \), we can express the equation of the line in point-slope form. The equation of the line can be written as:
\[ y - k = m(x - h) \]
Rearranging gives:
\[ y = mx - mh + k \]

### Step 3: Find the Intersection Points
To find the points where this line intersects the hyperbola, we substitute \( y \) from the line equation into the hyperbola equation:
\[ x(mx - mh + k) = 1 \]
This simplifies to:
\[ mx^2 - (mh - k)x - 1 = 0 \]

### Step 4: Use the Midpoint Formula
The roots of this quadratic equation represent the x-coordinates of the points where the line intersects the hyperbola. Let the roots be \( x_1 \) and \( x_2 \). The midpoint \( h \) of these roots can be expressed as:
\[ h = \frac{x_1 + x_2}{2} \]
Using Vieta's formulas, we know that:
\[ x_1 + x_2 = -\frac{b}{a} = \frac{mh - k}{m} \]
Thus:
\[ h = \frac{1}{2m}(mh - k) \]

### Step 5: Solve for \( k \)
From the midpoint formula, we can express \( k \) in terms of \( h \):
\[ k = -mh \]

### Step 6: Substitute and Find the Locus
Now, substituting \( k = -mh \) back into the equation gives us:
\[ y = -mx \]
This is the equation of a line through the origin with slope \( -m \).

### Final Result
The locus of the midpoint of the line segment is:
\[ y + mx = 0 \]