To solve the problem, we need to find the area of triangle \( PF_1F_2 \) where \( P \) is a point on the ellipse defined by the equation
\[
\frac{x^2}{9} + \frac{y^2}{4} = 1
\]
and the relationship \( |PF_1| = 2|PF_2| \).
### Step 1: Identify the parameters of the ellipse
The given ellipse can be rewritten in standard form as:
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\]
where \( a^2 = 9 \) and \( b^2 = 4 \). Thus, we have:
\[
a = 3 \quad \text{and} \quad b = 2
\]
### Step 2: Find the foci of the ellipse
The foci \( F_1 \) and \( F_2 \) of the ellipse are located at \( (ae, 0) \) and \( (-ae, 0) \) respectively, 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:
\[
F_1 = (3e, 0) = \left(3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\sqrt{5}, 0)
\]
\[
F_2 = (-3e, 0) = \left(-3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (-\sqrt{5}, 0)
\]
### Step 3: Set up the equation based on the condition
According to the problem, we have:
\[
|PF_1| = 2|PF_2|
\]
Let \( |PF_2| = d \). Then, \( |PF_1| = 2d \). By the property of ellipses, we know that:
\[
|PF_1| + |PF_2| = 2a = 6
\]
Substituting the values we have:
\[
2d + d = 6 \implies 3d = 6 \implies d = 2
\]
Thus, we have:
\[
|PF_2| = 2 \quad \text{and} \quad |PF_1| = 4
\]
### Step 4: Find the coordinates of point \( P \)
Let the coordinates of point \( P \) be \( (x, y) \). The distances from \( P \) to the foci are given by:
\[
PF_1 = \sqrt{(x - \sqrt{5})^2 + y^2} = 4
\]
\[
PF_2 = \sqrt{(x + \sqrt{5})^2 + y^2} = 2
\]
Squaring both equations, we get:
1. \( (x - \sqrt{5})^2 + y^2 = 16 \)
2. \( (x + \sqrt{5})^2 + y^2 = 4 \)
### Step 5: Solve the system of equations
From the first equation:
\[
(x - \sqrt{5})^2 + y^2 = 16 \implies x^2 - 2x\sqrt{5} + 5 + y^2 = 16 \implies x^2 + y^2 - 2x\sqrt{5} = 11 \quad \text{(1)}
\]
From the second equation:
\[
(x + \sqrt{5})^2 + y^2 = 4 \implies x^2 + 2x\sqrt{5} + 5 + y^2 = 4 \implies x^2 + y^2 + 2x\sqrt{5} = -1 \quad \text{(2)}
\]
### Step 6: Eliminate \( y^2 \)
Subtract equation (2) from (1):
\[
(-2x\sqrt{5} - 2x\sqrt{5}) = 11 + 1 \implies -4x\sqrt{5} = 12 \implies x = -\frac{3}{\sqrt{5}} = -\frac{3\sqrt{5}}{5}
\]
Substituting \( x \) back into either equation to find \( y^2 \):
Using equation (1):
\[
\left(-\frac{3\sqrt{5}}{5}\right)^2 + y^2 - 2\left(-\frac{3\sqrt{5}}{5}\right)\sqrt{5} = 11
\]
Calculating \( y^2 \) gives us the coordinates of point \( P \).
### Step 7: Calculate the area of triangle \( PF_1F_2 \)
Using the formula for the area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates of \( P \), \( F_1 \), and \( F_2 \) will yield the area.
### Final Answer
After performing the calculations, we find that the area of triangle \( PF_1F_2 \) is:
\[
\text{Area} = 4 \text{ square units}
\]
