If the distances from P to the points ` ( 3, 4 ) , ( - 3, 4 ) ` are in the ratio 3 : 2, then the locus of P is

Answer

To find the locus of point \( P \) such that the distances from \( P \) to the points \( (3, 4) \) and \( (-3, 4) \) are in the ratio \( 3:2 \), we can follow these steps:

### Step 1: Define the point \( P \)
Let the coordinates of point \( P \) be \( (x, y) \).

### Step 2: Use the distance formula
The distance from \( P \) to the point \( (3, 4) \) is given by:
\[ d_1 = \sqrt{(x - 3)^2 + (y - 4)^2} \]
The distance from \( P \) to the point \( (-3, 4) \) is given by:
\[ d_2 = \sqrt{(x + 3)^2 + (y - 4)^2} \]

### Step 3: Set up the ratio of distances
According to the problem, the ratio of these distances is:
\[ \frac{d_1}{d_2} = \frac{3}{2} \]
Substituting the distances we found:
\[ \frac{\sqrt{(x - 3)^2 + (y - 4)^2}}{\sqrt{(x + 3)^2 + (y - 4)^2}} = \frac{3}{2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ \frac{(x - 3)^2 + (y - 4)^2}{(x + 3)^2 + (y - 4)^2} = \frac{9}{4} \]

### Step 5: Cross-multiply to simplify
Cross-multiplying results in:
\[ 4 \left( (x - 3)^2 + (y - 4)^2 \right) = 9 \left( (x + 3)^2 + (y - 4)^2 \right) \]

### Step 6: Expand both sides
Expanding both sides:
- Left side:
\[ 4 \left( (x - 3)^2 + (y - 4)^2 \right) = 4 \left( x^2 - 6x + 9 + y^2 - 8y + 16 \right) = 4x^2 - 24x + 100 - 32y \]
- Right side:
\[ 9 \left( (x + 3)^2 + (y - 4)^2 \right) = 9 \left( x^2 + 6x + 9 + y^2 - 8y + 16 \right) = 9x^2 + 54x + 225 - 72y \]

### Step 7: Set the equation to zero
Setting both sides equal gives:
\[ 4x^2 - 24x + 100 - 32y = 9x^2 + 54x + 225 - 72y \]
Rearranging this leads to:
\[ -5x^2 - 78x + 40y + 125 = 0 \]

### Step 8: Simplify the equation
Dividing through by \(-5\) gives:
\[ x^2 + \frac{78}{5}x - 8y - 25 = 0 \]

### Conclusion
The locus of point \( P \) is represented by the equation:
\[ 5x^2 + 5y^2 + 78x - 40y + 125 = 0 \]