Let `F_1`(0, 3) and `F_2`(0, -3) be two points and P be any point on the ellipse `25x^2 + 16y^2= 400`, then value of `(PF_1+ PF_2)` is

Answer

To solve the problem, we need to find the value of \( PF_1 + PF_2 \) for a point \( P \) on the ellipse defined by the equation \( 25x^2 + 16y^2 = 400 \).

### Step-by-Step Solution:

1. **Identify the Ellipse Equation:**
The given ellipse equation is:
\[ 25x^2 + 16y^2 = 400 \]

2. **Convert to Standard Form:**
To convert this to the standard form of an ellipse, we divide the entire equation by 400:
\[ \frac{25x^2}{400} + \frac{16y^2}{400} = 1 \]
Simplifying this gives:
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]

3. **Identify Values of \( a \) and \( b \):**
From the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can identify:
- \( a^2 = 16 \) which gives \( a = 4 \)
- \( b^2 = 25 \) which gives \( b = 5 \)

4. **Determine the Foci of the Ellipse:**
The foci of an ellipse are located at \( (0, \pm c) \), where \( c = \sqrt{b^2 - a^2} \).
\[ c = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Thus, the foci are at \( F_1(0, 3) \) and \( F_2(0, -3) \).

5. **Use the Property of Ellipses:**
A key property of ellipses is that for any point \( P \) on the ellipse, the sum of the distances from \( P \) to the foci \( F_1 \) and \( F_2 \) is constant and equal to the length of the major axis, which is \( 2b \).
\[ \text{Length of major axis} = 2b = 2 \times 5 = 10 \]

6. **Conclusion:**
Therefore, for any point \( P \) on the ellipse, the value of \( PF_1 + PF_2 \) is:
\[ PF_1 + PF_2 = 10 \]

### Final Answer:
The value of \( PF_1 + PF_2 \) is \( 10 \) units.

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