Find the locus of a point P which moves so that its distance from a given poin O is always in a given ratio (n : 1) to its distance from another given point A

Answer

To find the locus of a point P that moves such that its distance from a given point O is always in a given ratio (n:1) to its distance from another given point A, we can follow these steps:

### Step 1: Define the points and distances
Let the coordinates of point O be (x₀, y₀) and the coordinates of point A be (x₁, y₁). Let the coordinates of point P be (x, y). The distance from P to O is given by:
\[ d(P, O) = \sqrt{(x - x₀)² + (y - y₀)²} \]
The distance from P to A is given by:
\[ d(P, A) = \sqrt{(x - x₁)² + (y - y₁)²} \]

### Step 2: Set up the ratio
According to the problem, the distance from P to O is in the ratio n:1 to the distance from P to A. This can be expressed mathematically as:
\[ d(P, O) = n \cdot d(P, A) \]

### Step 3: Substitute the distances into the equation
Substituting the expressions for the distances into the ratio gives us:
\[ \sqrt{(x - x₀)² + (y - y₀)²} = n \cdot \sqrt{(x - x₁)² + (y - y₁)²} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides results in:
\[ (x - x₀)² + (y - y₀)² = n² \cdot ((x - x₁)² + (y - y₁)²) \]

### Step 5: Expand both sides
Expanding both sides gives:
\[ (x² - 2xx₀ + x₀² + y² - 2yy₀ + y₀²) = n²(x² - 2xx₁ + x₁² + y² - 2yy₁ + y₁²) \]

### Step 6: Rearrange the equation
Rearranging the equation to group like terms results in:
\[ (1 - n²)x² + (1 - n²)y² - 2x(1 - n²)x₀ - 2y(1 - n²)y₀ + (x₀² + y₀² - n²(x₁² + y₁²)) = 0 \]

### Step 7: Identify the locus
The resulting equation represents a conic section (specifically, a circle or an ellipse) depending on the value of n.

### Conclusion
Thus, the locus of point P is a conic section, which can be specifically identified as a circle or an ellipse based on the ratio n.

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