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

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given equation of the ellipse is:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
From this equation, we can identify:
- \( a^2 = 25 \) (thus \( a = 5 \))
- \( b^2 = 16 \) (thus \( b = 4 \))

### Step 2: Find the foci of the ellipse
The foci of an ellipse are given by the formula:
\[ c = \sqrt{a^2 - b^2} \]
Calculating \( c \):
\[ c = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Thus, the coordinates of the foci are:
\[ F_1(-3, 0) \quad \text{and} \quad F_2(3, 0) \]

### Step 3: Parametrize the point P on the ellipse
The coordinates of 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 \( PF_1F_2 \)
The area \( A \) of triangle \( PF_1F_2 \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( F_1F_2 \) is the distance between the foci:
\[ \text{base} = |F_2 - F_1| = |3 - (-3)| = 6 \]
The height is the y-coordinate of point \( P \), which is \( 4 \sin \theta \):
\[ A = \frac{1}{2} \times 6 \times 4 \sin \theta = 12 \sin \theta \]

### Step 5: Determine when the area is maximum
The area \( A = 12 \sin \theta \) is maximized when \( \sin \theta \) is maximum. The maximum value of \( \sin \theta \) is 1, which occurs at:
\[ \theta = \frac{\pi}{2} \]
Thus, the maximum area is:
\[ A_{\text{max}} = 12 \times 1 = 12 \]

### Step 6: Find the coordinates of point P when area is maximum
When \( \theta = \frac{\pi}{2} \):
\[ P(5 \cos \frac{\pi}{2}, 4 \sin \frac{\pi}{2}) = P(0, 4) \]

### Summary of Results
- The coordinates of the foci are \( F_1(-3, 0) \) and \( F_2(3, 0) \).
- The area of triangle \( PF_1F_2 \) is given by \( 12 \sin \theta \).
- The maximum area of the triangle is 12, occurring when \( \theta = \frac{\pi}{2} \), and the coordinates of point \( P \) at this maximum area are \( (0, 4) \).