Let P be a variable point on the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` with foci `F_1" and "F_2`. If A is the area of the `trianglePF_1F_2`, then the maximum value of A is

Answer

To find the maximum area of triangle \( PF_1F_2 \) where \( P \) is a variable point on the ellipse given by the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

we can follow these steps:

### Step 1: Identify 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 of the ellipse given by

\[ e = \sqrt{1 - \frac{b^2}{a^2}}. \]

### Step 2: Parameterize the point \( P \)
We can express the coordinates of point \( P \) on the ellipse using the parametric equations:

\[ P(a \cos \theta, b \sin \theta). \]

### Step 3: Calculate the area of triangle \( PF_1F_2 \)
The area \( A \) of triangle \( PF_1F_2 \) can be calculated using the formula for the area of a triangle given by vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]

Substituting the coordinates of \( P \), \( F_1 \), and \( F_2 \):

- \( P = (a \cos \theta, b \sin \theta) \)
- \( F_1 = (-ae, 0) \)
- \( F_2 = (ae, 0) \)

The area becomes:

\[ A = \frac{1}{2} \left| a \cos \theta (0 - 0) + (-ae)(0 - b \sin \theta) + (ae)(b \sin \theta - 0) \right|. \]

This simplifies to:

\[ A = \frac{1}{2} \left| -ae(-b \sin \theta) + ae(b \sin \theta) \right| = \frac{1}{2} \left| 2aeb \sin \theta \right| = aeb \sin \theta. \]

### Step 4: Maximize the area
To find the maximum area, we note that \( \sin \theta \) achieves its maximum value of 1. Thus,

\[ A_{\text{max}} = aeb. \]

### Step 5: Substitute the value of \( e \)
Substituting \( e = \sqrt{1 - \frac{b^2}{a^2}} \):

\[ A_{\text{max}} = ab \sqrt{1 - \frac{b^2}{a^2}}. \]

### Final Answer
The maximum area of triangle \( PF_1F_2 \) is:

\[ A_{\text{max}} = b \sqrt{a^2 - b^2}. \]

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