Let P be a variable point on the ellipse `x^(2)/25 + y^(2)/16 = 1` with foci at S and S'. If A be the area of triangle PSS' then the maximum value of A, is

Options

A. 12 sq units
B. 24 sq units
C. 36 sq units
D. 48 sq units

Answer

To find the maximum area of triangle PSS' formed by a variable point P on the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) with foci S and S', we can follow these steps:

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

### Step 2: Determine the foci of the ellipse
The foci of the ellipse are located at \( (c, 0) \) and \( (-c, 0) \), where \( c = \sqrt{a^2 - b^2} \).
Calculating \( c \):
\[ c = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Thus, the foci S and S' are at \( (3, 0) \) and \( (-3, 0) \).

### Step 3: Calculate the length of the segment SS'
The distance between the foci S and S' is given by:
\[ SS' = 2c = 2 \times 3 = 6 \]

### Step 4: Express the area of triangle PSS'
The area \( A \) of triangle PSS' can be expressed as:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( SS' = 6 \) and the height is the perpendicular distance from point P to the line joining S and S'.

### Step 5: Determine the maximum height
The height of the triangle is maximized when point P is at the topmost point of the ellipse, which occurs when \( P \) is on the y-axis. The coordinates of point P can be expressed as \( (0, y) \) where \( y \) is the maximum value of \( y \) on the ellipse.

From the ellipse equation:
\[ \frac{0^2}{25} + \frac{y^2}{16} = 1 \implies y^2 = 16 \implies y = 4 \text{ (maximum height)} \]

### Step 6: Calculate the maximum area
Now substituting the values into the area formula:
\[ A = \frac{1}{2} \times SS' \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12 \]

### Conclusion
The maximum area \( A \) of triangle PSS' is:
\[ \boxed{12} \]