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. 24 sq. units
B. 12 sq. units
C. 36 sq. units
D. none of these

Answer

To find the maximum area \( A \) of triangle \( PSS' \) where \( P \) is a variable point on the ellipse given by the equation

\[ \frac{x^2}{25} + \frac{y^2}{16} = 1, \]

we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be expressed in standard form where \( a^2 = 25 \) and \( b^2 = 16 \). Thus, we have:

\[ a = 5 \quad \text{and} \quad b = 4. \]

### Step 2: Calculate the foci of the ellipse
The foci \( S \) and \( S' \) of the ellipse are located at \( (\pm ae, 0) \), where \( e \) is the eccentricity. The eccentricity \( e \) is calculated as follows:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}. \]

Thus, the coordinates of the foci are:

\[ S(3, 0) \quad \text{and} \quad S'(-3, 0). \]

### Step 3: Determine 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 \( SS' \) is the distance between the foci:

\[ \text{base} = |S - S'| = |3 - (-3)| = 6. \]

The height of the triangle is the perpendicular distance from point \( P \) to the line segment \( SS' \).

### Step 4: Maximize the height
To maximize the area \( A \), we need to maximize the height, which corresponds to the \( y \)-coordinate of point \( P \). The maximum value of \( y \) on the ellipse occurs at the endpoints of the minor axis, which is at \( (0, b) \) or \( (0, 4) \).

### Step 5: Calculate the maximum area
Substituting the maximum height into the area formula:

\[ A = \frac{1}{2} \times 6 \times 4 = 12. \]

Thus, the maximum area \( A \) of triangle \( PSS' \) is:

\[ \boxed{12}. \]

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