P is any variable point on the ellipse `4x^(2) + 9y^(2) = 36` and `F_(1), F_(2)` are its foci. Maxium area of `trianglePF_(1)F_(2)` ( e is eccentricity of ellipse )

Options

A. 9e
B. 4e
C. 6e
D. 10e

Answer

To solve the problem of finding the maximum area of triangle \( PF_1F_2 \) where \( P \) is a point on the ellipse \( 4x^2 + 9y^2 = 36 \) and \( F_1, F_2 \) are its foci, we can follow these steps:

### Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is:
\[ 4x^2 + 9y^2 = 36 \]
Dividing the entire equation by 36, we get:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This shows that \( a^2 = 9 \) and \( b^2 = 4 \).

### Step 2: Identify the semi-major and semi-minor axes
From the equation, we can determine:
\[ a = 3 \quad (a \text{ is the semi-major axis}) \]
\[ b = 2 \quad (b \text{ is the semi-minor axis}) \]

### Step 3: Calculate the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Determine the foci of the ellipse
The foci \( F_1 \) and \( F_2 \) are located at:
\[ F_1 = (-ae, 0) = \left(-3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (-\sqrt{5}, 0) \]
\[ F_2 = (ae, 0) = \left(3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\sqrt{5}, 0) \]

### Step 5: Parametrize the point \( P \) on the ellipse
The coordinates of point \( P \) on the ellipse can be expressed parametrically as:
\[ P = (a \cos \theta, b \sin \theta) = (3 \cos \theta, 2 \sin \theta) \]

### Step 6: Calculate the area of triangle \( PF_1F_2 \)
The area \( A \) of triangle \( PF_1F_2 \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the coordinates:
- \( P(3 \cos \theta, 2 \sin \theta) \)
- \( F_1(-\sqrt{5}, 0) \)
- \( F_2(\sqrt{5}, 0) \)

We have:
\[ A = \frac{1}{2} \left| 3 \cos \theta (0 - 0) + (-\sqrt{5})(0 - 2 \sin \theta) + \sqrt{5}(2 \sin \theta - 0) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 2\sqrt{5} \sin \theta + 2\sqrt{5} \sin \theta \right| = \frac{1}{2} \left| 4\sqrt{5} \sin \theta \right| = 2\sqrt{5} |\sin \theta| \]

### Step 7: Find the maximum area
The maximum value of \( |\sin \theta| \) is 1, which occurs when \( \theta = \frac{\pi}{2} \) or \( \theta = \frac{3\pi}{2} \). Thus, the maximum area is:
\[ A_{\text{max}} = 2\sqrt{5} \cdot 1 = 2\sqrt{5} \]

### Step 8: Express the area in terms of eccentricity
Since \( e = \frac{\sqrt{5}}{3} \), we can express the maximum area in terms of \( e \):
\[ A_{\text{max}} = 6e \]

### Final Answer
The maximum area of triangle \( PF_1F_2 \) is:
\[ \boxed{6e} \]