Let P be a variable on the ellipse `(x^(2))/(25)+ (y^(2))/(16) =1` with foci at `F_(1) and F_(2)`

Options

A. Area of `trianglePF_(1)F_(2)` is `12 sin theta`
B. Area of `triantglePF_(1)F_(2)` is maximum when `theta = (pi)/(2)`
C. Co-ordinate of P are (0, 4)
D. Centre of the ellipse is (1, 2)

Answer

To solve the problem, we need to analyze the given ellipse and the variable point \( P \) on it. The equation of the ellipse is:

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

### Step 1: Identify the parameters of the ellipse
The 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 \( F_1 \) and \( F_2 \) 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 coordinates of the foci are:
- \( F_1 = (3, 0) \)
- \( F_2 = (-3, 0) \)

### Step 3: Parametrize the variable point \( P \)
A point \( P \) on the ellipse can be expressed in parametric form as:
\[ P = (5 \cos \theta, 4 \sin \theta) \]

### Step 4: Calculate the area of triangle \( F_1PF_2 \)
The area \( A \) of triangle \( F_1PF_2 \) 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 \( F_1 \) and \( F_2 \), which is:
\[ \text{base} = |F_1 - F_2| = |3 - (-3)| = 6 \]

The height of the triangle is the y-coordinate of point \( P \), which is \( 4 \sin \theta \).

Thus, the area becomes:
\[ A = \frac{1}{2} \times 6 \times (4 \sin \theta) = 12 \sin \theta \]

### Step 5: Find the maximum area
To find the maximum area, we need to maximize \( 12 \sin \theta \). The maximum value of \( \sin \theta \) is 1, which occurs when \( \theta = \frac{\pi}{2} \).

Substituting \( \sin \theta = 1 \):
\[ A_{\text{max}} = 12 \times 1 = 12 \]

### Step 6: Determine the coordinates of point \( P \) at maximum area
When \( \sin \theta = 1 \), we have:
\[ \theta = \frac{\pi}{2} \implies P = (5 \cos \theta, 4 \sin \theta) = (5 \cdot 0, 4 \cdot 1) = (0, 4) \]

### Conclusion
The maximum area of triangle \( F_1PF_2 \) is 12, and it occurs when the coordinates of point \( P \) are \( (0, 4) \).

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