A variable straight line of slope `4` intersects the hyperbola `xy=1` at two points. The locus of the point which divides the line segment between these two points in the ratio `1 : 2` is

Answer

To solve the problem step by step, we start with the given information and work through the necessary calculations to find the locus of the point that divides the line segment between the two intersection points of the line and the hyperbola.

### Step 1: Equation of the Line
Given that the slope of the line is 4, we can write the equation of the line in slope-intercept form:
\[ y = 4x + c \]
where \( c \) is a constant.

### Step 2: Intersection with the Hyperbola
The hyperbola is given by the equation:
\[ xy = 1 \]
To find the points of intersection, we substitute \( y \) from the line equation into the hyperbola's equation:
\[ x(4x + c) = 1 \]
This simplifies to:
\[ 4x^2 + cx - 1 = 0 \]
This is a quadratic equation in \( x \).

### Step 3: Finding the Points of Intersection
Let the roots of the quadratic equation be \( t_1 \) and \( t_2 \). According to Vieta's formulas, we have:
- Sum of roots: \( t_1 + t_2 = -\frac{c}{4} \)
- Product of roots: \( t_1 t_2 = -\frac{1}{4} \)

### Step 4: Coordinates of Intersection Points
The coordinates of the intersection points can be expressed as:
- Point A: \( (t_1, \frac{1}{t_1}) \)
- Point B: \( (t_2, \frac{1}{t_2}) \)

### Step 5: Finding the Point that Divides the Segment
Let \( P(h, k) \) be the point that divides the line segment AB in the ratio \( 1:2 \). Using the section formula, the coordinates of point \( P \) can be calculated as:
\[ h = \frac{2t_1 + t_2}{3}, \quad k = \frac{2\left(\frac{1}{t_1}\right) + \left(\frac{1}{t_2}\right)}{3} \]
Substituting for \( k \):
\[ k = \frac{2/t_1 + 1/t_2}{3} = \frac{2t_2 + t_1}{3t_1t_2} \]

### Step 6: Expressing \( t_1 \) and \( t_2 \) in terms of \( h \) and \( k \)
From the equations:
1. \( t_1 + t_2 = -\frac{c}{4} \)
2. \( t_1 t_2 = -\frac{1}{4} \)

We can express \( t_1 \) and \( t_2 \) in terms of \( h \) and \( k \):
- From the first equation, we can express \( c \) as \( c = -4(t_1 + t_2) \).
- From the second equation, we can substitute \( t_2 = -\frac{1}{4t_1} \) into \( t_1 + t_2 \) to find a relationship.

### Step 7: Substitute and Simplify
Substituting the expressions for \( t_1 \) and \( t_2 \) into the equations for \( h \) and \( k \) gives us a relationship between \( h \) and \( k \). After simplification, we find:
\[ 16h^2 + k^2 + 10hk = 2 \]

### Final Answer
Thus, the locus of the point \( P(h, k) \) is given by:
\[ 16h^2 + k^2 + 10hk = 2 \]