A variable straight line drawn through the point of intersection of lines `x/a+y/b=1` and `x/b+y/a=1` meets the co ordinates axes in A and B. Locus of the mid point of AB is ……….

Answer

To solve the problem step by step, we will first find the point of intersection of the given lines, then derive the equation of the variable line, find the coordinates of points A and B where the line intersects the axes, and finally determine the locus of the midpoint of AB.

### Step 1: Find the Point of Intersection of the Lines
The two lines given are:
1. \( \frac{x}{a} + \frac{y}{b} = 1 \)
2. \( \frac{x}{b} + \frac{y}{a} = 1 \)

To find the intersection, we can express \( y \) in terms of \( x \) from both equations.

From the first equation:
\[ y = b \left(1 - \frac{x}{a}\right) = b - \frac{bx}{a} \]

From the second equation:
\[ y = a \left(1 - \frac{x}{b}\right) = a - \frac{ax}{b} \]

Setting both expressions for \( y \) equal to each other:
\[ b - \frac{bx}{a} = a - \frac{ax}{b} \]

Cross-multiplying to eliminate the fractions:
\[ b^2 - b \cdot \frac{bx}{a} = a^2 - a \cdot \frac{ax}{b} \]
\[ b^2 - \frac{b^2x}{a} = a^2 - \frac{a^2x}{b} \]

Rearranging gives:
\[ b^2 + \frac{a^2x}{b} - \frac{b^2x}{a} = a^2 \]

Now, we can solve for \( x \) and then substitute back to find \( y \).

### Step 2: Derive the Equation of the Variable Line
The variable line passing through the intersection point can be expressed in the form:
\[ \frac{x}{a} + \frac{y}{b} - 1 + \lambda \left(\frac{x}{b} + \frac{y}{a} - 1\right) = 0 \]

This simplifies to:
\[ \left(1 + \lambda\right)\frac{x}{a} + \left(1 + \lambda\right)\frac{y}{b} - (1 + \lambda) = 0 \]

### Step 3: Find the Coordinates of Points A and B
To find the points A and B where the line intersects the axes, we set \( y = 0 \) for point A and \( x = 0 \) for point B.

1. For point A (\( y = 0 \)):
\[ \frac{x}{a} + \lambda \frac{x}{b} = 1 \implies x\left(\frac{1}{a} + \lambda \frac{1}{b}\right) = 1 \implies x = \frac{1}{\frac{1}{a} + \lambda \frac{1}{b}} = \frac{ab}{b + \lambda a} \]

2. For point B (\( x = 0 \)):
\[ \frac{y}{b} + \lambda \frac{y}{a} = 1 \implies y\left(\frac{1}{b} + \lambda \frac{1}{a}\right) = 1 \implies y = \frac{1}{\frac{1}{b} + \lambda \frac{1}{a}} = \frac{ab}{a + \lambda b} \]

### Step 4: Find the Midpoint of AB
The midpoint \( P(h, k) \) of points A and B is given by:
\[ h = \frac{x_A + x_B}{2} = \frac{\frac{ab}{b + \lambda a} + 0}{2} = \frac{ab}{2(b + \lambda a)} \]
\[ k = \frac{y_A + y_B}{2} = \frac{0 + \frac{ab}{a + \lambda b}}{2} = \frac{ab}{2(a + \lambda b)} \]

### Step 5: Locus of the Midpoint P
To find the locus, we express \( h \) and \( k \) in terms of \( \lambda \):
\[ \frac{1}{2h} = \frac{b + \lambda a}{ab} \quad \text{and} \quad \frac{1}{2k} = \frac{a + \lambda b}{ab} \]

Adding these two equations:
\[ \frac{1}{2h} + \frac{1}{2k} = \frac{b + \lambda a + a + \lambda b}{ab} = \frac{(a + b) + \lambda(a + b)}{ab} \]

This simplifies to:
\[ \frac{1}{2h} + \frac{1}{2k} = \frac{(a + b)(1 + \lambda)}{ab} \]

### Final Result
Thus, the locus of the midpoint \( P \) is given by:
\[ \frac{1}{2x} + \frac{1}{2y} = \frac{1}{a} + \frac{1}{b} \]