Consider an ellipse, whose centre is at the origin and its major axis is along x-axis. if its eccentricity is 3/5 and the distance between its foci is 6, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is

Options

A. 8
B. 32
C. 80
D. 40

Answer

To solve the problem step by step, we will start by determining the parameters of the ellipse and then use them to find the area of the quadrilateral inscribed in the ellipse.

### Step 1: Understand the parameters of the ellipse
Given:
- The center of the ellipse is at the origin (0,0).
- The major axis is along the x-axis.
- The eccentricity \( e = \frac{3}{5} \).
- The distance between the foci is 6.

### Step 2: Use the distance between the foci to find \( a \)
The distance between the foci is given by \( 2ae \). Since this distance is 6, we have:
\[ 2ae = 6 \implies ae = 3 \]

### Step 3: Substitute the value of \( e \) to find \( a \)
Substituting \( e = \frac{3}{5} \) into the equation \( ae = 3 \):
\[ a \left(\frac{3}{5}\right) = 3 \implies a = 3 \cdot \frac{5}{3} = 5 \]

### Step 4: Use the eccentricity to find \( b \)
We know the relationship between \( a \), \( b \), and \( e \):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides gives:
\[ \left(\frac{3}{5}\right)^2 = 1 - \frac{b^2}{5^2} \implies \frac{9}{25} = 1 - \frac{b^2}{25} \]
Rearranging gives:
\[ \frac{b^2}{25} = 1 - \frac{9}{25} = \frac{16}{25} \implies b^2 = 16 \implies b = 4 \]

### Step 5: Write the equation of the ellipse
The equation of the ellipse can now be written as:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 6: Identify the vertices of the ellipse
The vertices of the ellipse are:
- Along the x-axis: \( (5, 0) \) and \( (-5, 0) \)
- Along the y-axis: \( (0, 4) \) and \( (0, -4) \)

### Step 7: Calculate the area of the quadrilateral
The quadrilateral formed by these vertices is a rectangle with:
- Base = distance between the x-axis vertices = \( 5 - (-5) = 10 \)
- Height = distance between the y-axis vertices = \( 4 - (-4) = 8 \)

The area \( A \) of the rectangle can be calculated as:
\[ A = \text{Base} \times \text{Height} = 10 \times 8 = 80 \text{ square units} \]

### Final Answer
The area of the quadrilateral inscribed in the ellipse is \( 80 \) square units.

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