The ends of a rod of length l move on two mutually perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Answer

To find the locus of the point on a rod of length \( l \) that divides it in the ratio \( 1:2 \) while the ends of the rod move on two mutually perpendicular lines (the x-axis and the y-axis), we can follow these steps:

### Step 1: Define the Points
Let the ends of the rod be at points \( A(a, 0) \) on the x-axis and \( B(0, b) \) on the y-axis. The length of the rod \( AB \) is given by the distance formula:
\[ AB = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \]
Since the length of the rod is \( l \), we have:
\[ \sqrt{a^2 + b^2} = l \]
Squaring both sides gives:
\[ a^2 + b^2 = l^2 \quad \text{(1)} \]

### Step 2: Find the Coordinates of Point P
Let point \( P(h, k) \) divide the rod \( AB \) in the ratio \( 1:2 \). By the section formula, the coordinates of point \( P \) can be calculated as follows:
\[ h = \frac{m x_2 + n x_1}{m+n} = \frac{1 \cdot 0 + 2 \cdot a}{1 + 2} = \frac{2a}{3} \]
\[ k = \frac{m y_2 + n y_1}{m+n} = \frac{1 \cdot b + 2 \cdot 0}{1 + 2} = \frac{b}{3} \]

### Step 3: Express \( a \) and \( b \) in terms of \( h \) and \( k \)
From the equations for \( h \) and \( k \):
\[ a = \frac{3h}{2} \quad \text{(2)} \]
\[ b = 3k \quad \text{(3)} \]

### Step 4: Substitute \( a \) and \( b \) into Equation (1)
Substituting equations (2) and (3) into equation (1):
\[ \left(\frac{3h}{2}\right)^2 + (3k)^2 = l^2 \]
Expanding this gives:
\[ \frac{9h^2}{4} + 9k^2 = l^2 \]

### Step 5: Clear the Fraction
To eliminate the fraction, multiply through by 4:
\[ 9h^2 + 36k^2 = 4l^2 \]

### Step 6: Rewrite in Terms of x and y
Replacing \( h \) and \( k \) with \( x \) and \( y \) respectively, we get:
\[ 9x^2 + 36y^2 = 4l^2 \]

### Final Result
The locus of the point \( P \) is given by the equation:
\[ 9x^2 + 36y^2 = 4l^2 \]