A point P(x, y) moves so that the product of the slopes of the two lines joining P to the two points (-2, 1) and (6, 5) is -4. Show that the locus is an ellipse and locate its centre.

Answer

To solve the problem step by step, we need to find the locus of the point \( P(x, y) \) such that the product of the slopes of the lines joining \( P \) to the points \( A(-2, 1) \) and \( B(6, 5) \) is equal to \(-4\).

### Step 1: Find the slopes of the lines PA and PB

The slope of the line joining point \( P(x, y) \) to point \( A(-2, 1) \) is given by:
\[ m_{PA} = \frac{y - 1}{x + 2} \]
The slope of the line joining point \( P(x, y) \) to point \( B(6, 5) \) is given by:
\[ m_{PB} = \frac{y - 5}{x - 6} \]

### Step 2: Set up the equation for the product of slopes

According to the problem, the product of these slopes is:
\[ m_{PA} \cdot m_{PB} = -4 \]
Substituting the expressions for the slopes, we have:
\[ \left(\frac{y - 1}{x + 2}\right) \cdot \left(\frac{y - 5}{x - 6}\right) = -4 \]

### Step 3: Simplify the equation

Cross-multiplying gives:
\[ (y - 1)(y - 5) = -4 \cdot (x + 2)(x - 6) \]
Expanding both sides:
\[ y^2 - 6y + 5 = -4(x^2 - 4x - 12) \]
This simplifies to:
\[ y^2 - 6y + 5 = -4x^2 + 16x + 48 \]

### Step 4: Rearranging the equation

Rearranging gives:
\[ 4x^2 + y^2 - 6y - 16x + 5 - 48 = 0 \]
This simplifies to:
\[ 4x^2 + y^2 - 6y - 16x - 43 = 0 \]

### Step 5: Completing the square

Now we will complete the square for both \( x \) and \( y \).

For \( x \):
\[ 4(x^2 - 4x) = 4((x - 2)^2 - 4) = 4(x - 2)^2 - 16 \]

For \( y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the equation:
\[ 4((x - 2)^2 - 4) + (y - 3)^2 - 9 - 43 = 0 \]
This simplifies to:
\[ 4(x - 2)^2 + (y - 3)^2 - 16 - 9 - 43 = 0 \]
\[ 4(x - 2)^2 + (y - 3)^2 - 68 = 0 \]
\[ 4(x - 2)^2 + (y - 3)^2 = 68 \]

### Step 6: Divide by 68

Dividing the entire equation by 68 gives:
\[ \frac{(x - 2)^2}{17} + \frac{(y - 3)^2}{68} = 1 \]

### Conclusion: Identify the locus and center

This is the standard form of the equation of an ellipse. The center of the ellipse is at the point \( (2, 3) \).

### Final Answer:
The locus of the point \( P(x, y) \) is an ellipse, and its center is at \( (2, 3) \).

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