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 will follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
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 is the standard form of the ellipse where \( a^2 = 9 \) and \( b^2 = 4 \). Thus, \( a = 3 \) and \( b = 2 \).

### Step 2: Determine the foci of the ellipse
The foci \( F_1 \) and \( F_2 \) of the ellipse are located at:
\[ F_1 = (-ae, 0) \quad \text{and} \quad F_2 = (ae, 0) \]
where \( e \) is the eccentricity given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]
Thus, the coordinates of the foci are:
\[ F_1 = \left(-3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (-\sqrt{5}, 0) \quad \text{and} \quad F_2 = \left(3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\sqrt{5}, 0) \]

### Step 3: Parameterize the point \( P \) on the ellipse
A point \( P \) on the ellipse can be parameterized as:
\[ P = (3 \cos \theta, 2 \sin \theta) \]

### Step 4: Calculate the area of triangle \( PF_1F_2 \)
The area \( A \) of triangle \( PF_1F_2 \) can be calculated using the determinant 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 \( P = (3 \cos \theta, 2 \sin \theta) \), \( F_1 = (-\sqrt{5}, 0) \), and \( F_2 = (\sqrt{5}, 0) \):
\[ A = \frac{1}{2} \left| 3 \cos \theta (0 - 0) + (-\sqrt{5})(0 - 2 \sin \theta) + \sqrt{5}(2 \sin \theta - 0) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| -\sqrt{5} \cdot (-2 \sin \theta) + \sqrt{5} \cdot (2 \sin \theta) \right| = \frac{1}{2} \left| 2\sqrt{5} \sin \theta + 2\sqrt{5} \sin \theta \right| = \frac{1}{2} \cdot 4\sqrt{5} \sin \theta = 2\sqrt{5} \sin \theta \]

### Step 5: Maximize the area
To find the maximum area, we need to maximize \( 2\sqrt{5} \sin \theta \). The maximum value of \( \sin \theta \) is 1, which occurs when \( \theta = \frac{\pi}{2} \). Thus, the maximum area is:
\[ A_{\text{max}} = 2\sqrt{5} \cdot 1 = 2\sqrt{5} \]

### Final Result
The maximum area of triangle \( PF_1F_2 \) is:
\[ \boxed{2\sqrt{5}} \]