If P is a point on the ellipse of eccentricity e and A, A 1 are the vertices and S, S' are the foci then area of SPS' : area of APA' =

Options

A. `e^(3)`
B. `e^(2)`
C. e
D. `(1)/(e )`

Answer

To solve the problem, we need to find the ratio of the areas of two triangles: \( \triangle SPS' \) and \( \triangle APA' \), where \( P \) is a point on the ellipse, \( A \) and \( A' \) are the vertices, and \( S \) and \( S' \) are the foci.

### Step 1: Understand the coordinates of the points
- Let the vertices \( A \) and \( A' \) be at \( (a, 0) \) and \( (-a, 0) \) respectively.
- The foci \( S \) and \( S' \) are located at \( (ae, 0) \) and \( (-ae, 0) \) respectively.
- The point \( P \) on the ellipse can be represented as \( (x, y) \).

### Step 2: Calculate the area of triangle \( SPS' \)
- The base of triangle \( SPS' \) is the distance between the foci \( S \) and \( S' \), which is \( 2ae \).
- The height of the triangle from point \( P \) to the line joining \( S \) and \( S' \) is the y-coordinate of point \( P \), which is \( y \).
- The area \( A_1 \) of triangle \( SPS' \) can be calculated using the formula for the area of a triangle:
\[ A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2ae) \times y = ae \cdot y \]

### Step 3: Calculate the area of triangle \( APA' \)
- The base of triangle \( APA' \) is the distance between the vertices \( A \) and \( A' \), which is also \( 2a \).
- The height of this triangle is the same as before, which is \( y \).
- The area \( A_2 \) of triangle \( APA' \) can be calculated similarly:
\[ A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2a) \times y = a \cdot y \]

### Step 4: Find the ratio of the areas
- Now, we can find the ratio of the areas of the two triangles:
\[ \frac{A_1}{A_2} = \frac{ae \cdot y}{a \cdot y} \]
- Simplifying this gives:
\[ \frac{A_1}{A_2} = e \]

### Conclusion
Thus, the ratio of the areas \( \text{Area of } SPS' : \text{Area of } APA' = e : 1 \).