If `P(x ,y)` is any point on the ellipse `16 x^2+25 y^2=400` and `f_1=(3,0)F_2=(-3,0)` , then find the value of `P F_1+P F_2dot`

Options

A. 8
B. 6
C. 10
D. 12

Explanation

To solve the problem, we need to find the value of \( PF_1 + PF_2 \) where \( P(x, y) \) is any point on the ellipse defined by the equation \( 16x^2 + 25y^2 = 400 \), and the foci \( F_1 = (3, 0) \) and \( F_2 = (-3, 0) \).

### Step-by-Step Solution:

1. **Convert the Ellipse Equation to Standard Form:**
The given equation of the ellipse is:
\[ 16x^2 + 25y^2 = 400 \]
Divide the entire equation by 400:
\[ \frac{16x^2}{400} + \frac{25y^2}{400} = 1 \]
Simplifying gives:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
This is the standard form of an ellipse.

2. **Identify Parameters of the Ellipse:**
From the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we identify:
- \( a^2 = 25 \) so \( a = 5 \)
- \( b^2 = 16 \) so \( b = 4 \)

3. **Calculate the Eccentricity \( e \):**
The eccentricity \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

4. **Determine the Foci of the Ellipse:**
The foci \( F_1 \) and \( F_2 \) are located at:
\[ (\pm ae, 0) = (\pm 5 \cdot \frac{3}{5}, 0) = (\pm 3, 0) \]
Thus, \( F_1 = (3, 0) \) and \( F_2 = (-3, 0) \).

5. **Use the Property of the Ellipse:**
A fundamental property of ellipses states that for any point \( P \) on the ellipse, the sum of the distances from \( P \) to the foci is equal to the length of the major axis, which is \( 2a \):
\[ PF_1 + PF_2 = 2a = 2 \times 5 = 10 \]

### Final Answer:
Thus, the value of \( PF_1 + PF_2 \) is \( 10 \).

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