A series of concentric ellipses `E_1,E_2, E_3..., E_n` are drawn such that E touches the extremities of the major axis of `E_(n-1)`, and the foci of `E_n` coincide with the extremities of minor axis of `E_(n-1)` If the eccentricity of the ellipses is independent of n, then the value of the eccentricity, is

Options

A. (a)`(3-sqrt5)/2`
B. (b)`(sqrt5-1)/2`
C. (c)`(2-sqrt3)/(2)`
D. (d)`(sqrt3-1)/2`

Answer

To solve the problem step by step, we will analyze the properties of the concentric ellipses \( E_n \) and \( E_{n-1} \) and derive the eccentricity \( e \).

### Step 1: Understanding the Geometry of the Ellipses
We have a series of concentric ellipses \( E_1, E_2, \ldots, E_n \). The ellipse \( E_n \) touches the extremities of the major axis of \( E_{n-1} \) and its foci coincide with the extremities of the minor axis of \( E_{n-1} \).

### Step 2: Define the Parameters of the Ellipses
Let:
- The semi-major axis of \( E_{n-1} \) be \( a_{n-1} \).
- The semi-minor axis of \( E_{n-1} \) be \( b_{n-1} \).
- The eccentricity of the ellipses be \( e \).

The relationship between the semi-major and semi-minor axes and the eccentricity is given by:
\[ b_{n-1} = a_{n-1} \sqrt{1 - e^2} \]

### Step 3: Determine the Dimensions of \( E_n \)
Since \( E_n \) touches the extremities of the major axis of \( E_{n-1} \), the semi-major axis of \( E_n \) is equal to \( a_{n-1} \):
\[ a_n = a_{n-1} \]

The foci of \( E_n \) coincide with the extremities of the minor axis of \( E_{n-1} \), which means:
\[ c_n = b_{n-1} = a_{n-1} \sqrt{1 - e^2} \]

### Step 4: Relate the Semi-Major and Semi-Minor Axes of \( E_n \)
The semi-minor axis of \( E_n \) can be expressed in terms of its semi-major axis and eccentricity:
\[ b_n = a_n \sqrt{1 - e^2} = a_{n-1} \sqrt{1 - e^2} \]

### Step 5: Establish the Relationship Between \( E_n \) and \( E_{n-1} \)
From the above, we have:
\[ c_n = a_{n-1} \sqrt{1 - e^2} \]
And since \( c_n = a_n e \):
\[ a_{n-1} e = a_{n-1} \sqrt{1 - e^2} \]

### Step 6: Simplify the Equation
Dividing both sides by \( a_{n-1} \) (assuming \( a_{n-1} \neq 0 \)):
\[ e = \sqrt{1 - e^2} \]

### Step 7: Square Both Sides
Squaring both sides gives:
\[ e^2 = 1 - e^2 \]
\[ 2e^2 = 1 \]
\[ e^2 = \frac{1}{2} \]

### Step 8: Find the Value of Eccentricity
Taking the square root of both sides:
\[ e = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Conclusion
The eccentricity \( e \) of the ellipses is:
\[ \boxed{\frac{\sqrt{2}}{2}} \]