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 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 equation can be rewritten in standard form:
- \( a^2 = 25 \) which gives \( a = 5 \)
- \( b^2 = 16 \) which gives \( b = 4 \)

### Step 2: Calculate the eccentricity (e) of the ellipse
The formula for eccentricity \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Find the coordinates of the foci (S and S')
The foci of the ellipse are located at \( (ae, 0) \) and \( (-ae, 0) \):
\[ S = (3, 0) \quad \text{and} \quad S' = (-3, 0) \]

### Step 4: Area of triangle PSS'
The area \( A \) of triangle PSS' can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the distance between the foci \( SS' \), which is \( 6 \) (from \( -3 \) to \( 3 \)), and the height is the y-coordinate of point \( P \).

### Step 5: Express the height in terms of x
From the ellipse equation, we can express \( y \) in terms of \( x \):
\[ y = 4 \sqrt{1 - \frac{x^2}{25}} \]
Thus, the area becomes:
\[ A = \frac{1}{2} \times 6 \times y = 3y = 3 \times 4 \sqrt{1 - \frac{x^2}{25}} = 12 \sqrt{1 - \frac{x^2}{25}} \]

### Step 6: Maximize the area A
To find the maximum area, we need to maximize \( A = 12 \sqrt{1 - \frac{x^2}{25}} \). We can do this by finding the derivative and setting it to zero:
\[ \frac{dA}{dx} = 12 \cdot \frac{1}{2} \cdot (1 - \frac{x^2}{25})^{-1/2} \cdot \left(-\frac{2x}{25}\right) \]
Setting \( \frac{dA}{dx} = 0 \) gives:
\[ -\frac{12x}{25\sqrt{1 - \frac{x^2}{25}}} = 0 \]
This implies \( x = 0 \).

### Step 7: Calculate the maximum area
Substituting \( x = 0 \) back into the area formula:
\[ A_{\text{max}} = 12 \sqrt{1 - \frac{0^2}{25}} = 12 \times 1 = 12 \]

Thus, the maximum area of triangle PSS' is:
\[ \boxed{12} \]