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 of constant slope \( m \) that intersects the branches of the rectangular hyperbola \( xy = 1 \), we can follow these steps:

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

### Step 2: Equation of the Line with Constant Slope
Let the equation of the line with slope \( m \) be:
\[ y = mx + c \]
where \( c \) is the y-intercept.

### Step 3: Find Points of Intersection
To find the points where this line intersects the hyperbola, we substitute \( y \) from the line's equation into the hyperbola's equation:
\[ x(mx + c) = 1 \]
This simplifies to:
\[ mx^2 + cx - 1 = 0 \]
This is a quadratic equation in \( x \).

### Step 4: Use the Quadratic Formula
The roots of this quadratic equation will give us the x-coordinates of the points of intersection. Using the quadratic formula:
\[ x = \frac{-c \pm \sqrt{c^2 + 4m}}{2m} \]

### Step 5: Find the Midpoint
Let the two points of intersection be \( (x_1, y_1) \) and \( (x_2, y_2) \). The x-coordinates are \( x_1 \) and \( x_2 \), and the y-coordinates can be found using the line equation:
\[ y_1 = mx_1 + c, \quad y_2 = mx_2 + c \]
The midpoint \( (H, K) \) is given by:
\[ H = \frac{x_1 + x_2}{2}, \quad K = \frac{y_1 + y_2}{2} \]

### Step 6: Use the Sum and Product of Roots
From the properties of quadratic equations, we know:
\[ x_1 + x_2 = -\frac{c}{m}, \quad x_1 x_2 = -\frac{1}{m} \]
Thus, the midpoint \( H \) becomes:
\[ H = \frac{-\frac{c}{m}}{2} = -\frac{c}{2m} \]
And for \( K \):
\[ K = \frac{(mx_1 + c) + (mx_2 + c)}{2} = mH + c \]

### Step 7: Substitute for \( c \)
From the equation of the line, we can express \( c \) in terms of \( H \) and \( K \):
\[ K = mH + c \implies c = K - mH \]

### Step 8: Substitute Back into the Midpoint Equation
Now, substituting \( c \) back into the expression for \( H \):
\[ H = -\frac{K - mH}{2m} \]
Rearranging gives:
\[ 2mH = -K + mH \implies H + \frac{K}{2m} = 0 \implies K = -2mH \]

### Step 9: Final Locus Equation
Thus, the locus of the midpoint \( (H, K) \) is:
\[ K = -2mH \]
This represents a straight line with slope \( -2m \).

### Conclusion
The locus of the midpoint of the portion of a line of constant slope \( m \) between two branches of the rectangular hyperbola \( xy = 1 \) is a straight line given by the equation:
\[ y = -2mx \]