Perimeter of a triangle formed by any point on the ellipse
`(x^(2))/(25) + (y^(2))/(16) = 1` and its foci is

Options

A. 13
B. 14
C. 15
D. 16

Answer

To find the perimeter of the triangle formed by any point on the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) and its foci, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be expressed in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where:
- \(a^2 = 25\) implies \(a = 5\)
- \(b^2 = 16\) implies \(b = 4\)

### Step 2: Calculate the distance \(c\) from the center to the foci
The relationship between \(a\), \(b\), and \(c\) in an ellipse is given by the equation:
\[ c^2 = a^2 - b^2 \]
Substituting the values:
\[ c^2 = 25 - 16 = 9 \implies c = 3 \]

### Step 3: Determine the coordinates of the foci
The foci of the ellipse are located at \((\pm c, 0)\). Therefore, the coordinates of the foci are:
- \(F_1 = (3, 0)\)
- \(F_2 = (-3, 0)\)

### Step 4: Use the property of the ellipse
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 \(2a\):
\[ PF_1 + PF_2 = 2a = 2 \times 5 = 10 \]

### Step 5: Calculate the distance between the foci
The distance between the two foci \(F_1\) and \(F_2\) is:
\[ F_1F_2 = 2c = 2 \times 3 = 6 \]

### Step 6: Calculate the perimeter of the triangle
The perimeter \(P\) of the triangle formed by the point \(P\) and the foci \(F_1\) and \(F_2\) is given by:
\[ P = PF_1 + PF_2 + F_1F_2 \]
Substituting the known values:
\[ P = 10 + 6 = 16 \]

### Conclusion
Thus, the perimeter of the triangle formed by any point on the ellipse and its foci is \(16\).

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