Suppose S and S' are foci of the ellipse `(x^(2))/(25)+(y^(2))/(16)=1`. If P is a variable point on the ellipse and if `Delta` is the area (in sq. units) of the triangle PSS' then the maximum value of `Delta` is double of

Options

A. Minimum value of `(2a^(8)+2b^(4))/(a^(4)b^(2))AA a, b in R`
B. Minimum value of `(3a^(8)+3b^(4))/(a^(4)b^(2))AA a, b in R`
C. `(4a^(8)+4b^(4))/(a^(4)b^(2))AA a, b in R`
D. `(6a^(8)+6b^(4))/(a^(4)b^(2))AA a, b in R`

Answer

To solve the problem, we need to find the maximum area of the triangle formed by a point \( P \) on the ellipse and its foci \( S \) and \( S' \). The ellipse is given by the equation:

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

### Step 1: Identify the parameters of the ellipse
The standard form of the ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

From the equation, we can identify:
- \( a^2 = 25 \) → \( a = 5 \)
- \( b^2 = 16 \) → \( b = 4 \)

### Step 2: Calculate the foci of the ellipse
The distance of the foci from the center is given by:

\[ c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \]

Thus, the coordinates of the foci \( S \) and \( S' \) are:
- \( S(3, 0) \)
- \( S'(-3, 0) \)

### Step 3: Parametrize the point \( P \) on the ellipse
A point \( P \) on the ellipse can be represented in parametric form as:

\[ P(5 \cos \theta, 4 \sin \theta) \]

### Step 4: Calculate the area of triangle \( PSS' \)
The area \( \Delta \) of triangle \( PSS' \) can be calculated using the determinant formula for the area of a triangle formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of \( P \), \( S \), and \( S' \):

\[ \Delta = \frac{1}{2} \left| 5 \cos \theta (0 - 0) + 3(0 - 4 \sin \theta) + (-3)(4 \sin \theta - 0) \right| \]

This simplifies to:

\[ \Delta = \frac{1}{2} \left| -12 \sin \theta \right| = 6 |\sin \theta| \]

### Step 5: Find the maximum value of the area
The maximum value of \( |\sin \theta| \) is 1. Therefore, the maximum area \( \Delta \) becomes:

\[ \Delta_{\text{max}} = 6 \times 1 = 6 \text{ square units} \]

### Step 6: Determine the final answer
According to the problem, the maximum value of \( \Delta \) is double of some value. Therefore, we have:

\[ \Delta_{\text{max}} = 2 \times x \implies 6 = 2x \implies x = 3 \]

Thus, the answer is:

\[ \text{The maximum value of } \Delta \text{ is double of } 3. \]