Let `F_1,F_2` be two foci of the ellipse `x^2/(p^2+2)+y^2/(p^2+4)=1` . Let P be any point on the ellipse , the maximum possible value of `PF_1 . PF_2 -p^2` is

Answer

To solve the problem, we need to find the maximum possible value of \( PF_1 \cdot PF_2 - p^2 \) for a point \( P \) on the ellipse defined by the equation:

\[ \frac{x^2}{p^2 + 2} + \frac{y^2}{p^2 + 4} = 1 \]

### Step 1: Identify the semi-major and semi-minor axes of the ellipse
The standard form of the ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

From the given equation, we can identify:

- \( a^2 = p^2 + 2 \)
- \( b^2 = p^2 + 4 \)

Since \( b^2 > a^2 \), we have:

- \( a = \sqrt{p^2 + 2} \)
- \( b = \sqrt{p^2 + 4} \)

### Step 2: Calculate the distance between the foci
The distance between the foci \( F_1 \) and \( F_2 \) of the ellipse is given by:

\[ c = \sqrt{b^2 - a^2} \]

Calculating \( c \):

\[ c = \sqrt{(p^2 + 4) - (p^2 + 2)} = \sqrt{2} \]

Thus, the coordinates of the foci are:

- \( F_1 = (-c, 0) = (-\sqrt{2}, 0) \)
- \( F_2 = (c, 0) = (\sqrt{2}, 0) \)

### Step 3: Use the property of the ellipse
For any point \( P \) on the ellipse, the sum of the distances to the foci is constant:

\[ PF_1 + PF_2 = 2b = 2\sqrt{p^2 + 4} \]

### Step 4: Apply the AM-GM inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:

\[ \frac{PF_1 + PF_2}{2} \geq \sqrt{PF_1 \cdot PF_2} \]

Substituting the known value:

\[ \sqrt{p^2 + 4} \geq \sqrt{PF_1 \cdot PF_2} \]

Squaring both sides gives:

\[ p^2 + 4 \geq PF_1 \cdot PF_2 \]

### Step 5: Rearranging the inequality
Rearranging the above inequality, we have:

\[ PF_1 \cdot PF_2 \leq p^2 + 4 \]

### Step 6: Find the maximum value of \( PF_1 \cdot PF_2 - p^2 \)
Thus, we can write:

\[ PF_1 \cdot PF_2 - p^2 \leq 4 \]

This means the maximum possible value of \( PF_1 \cdot PF_2 - p^2 \) is:

\[ \text{Maximum value} = 4 \]

### Conclusion
The maximum possible value of \( PF_1 \cdot PF_2 - p^2 \) is:

\[ \boxed{4} \]