A series of concentric ellipse `E_1,E_2,E_3,…,E_n` is constructed as follows: Ellipse `E_n` touches the extremities of the major axis of `E_(n-1)` and have its focii at the extremities of the minor axis of `E_(n-1)`. If eccentricity of ellipse `E_n` is `e_n`, then the locus of `(e_(n)^(2),e_(n-1)^(2))` is

Options

A. a parabola
B. an ellipse
C. a hyperbola
D. a rectangular hyperbola

Answer

To solve the problem step by step, we will analyze the given conditions and derive the required locus of the eccentricities of the ellipses.

### Step 1: Define the Ellipses
Let the ellipse \( E_n \) be defined by the equation:
\[ \frac{x^2}{a_n^2} + \frac{y^2}{b_n^2} = 1 \]
where \( a_n > b_n \). The eccentricity \( e_n \) of the ellipse is given by:
\[ e_n = \sqrt{1 - \frac{b_n^2}{a_n^2}} \]
From this, we can express \( b_n^2 \) in terms of \( e_n \):
\[ b_n^2 = a_n^2(1 - e_n^2) \tag{1} \]

### Step 2: Conditions for Ellipse \( E_n \)
According to the problem, the ellipse \( E_n \) touches the extremities of the major axis of \( E_{n-1} \) and has its foci at the extremities of the minor axis of \( E_{n-1} \). This gives us two key relationships:
1. The semi-minor axis of \( E_n \) is equal to the semi-minor axis of \( E_{n-1} \):
\[ b_n = b_{n-1} \tag{2} \]
2. The semi-major axis of \( E_{n-1} \) is given by:
\[ a_{n-1} = a_n e_n \tag{3} \]

### Step 3: Express \( b_{n-1}^2 \)
For the ellipse \( E_{n-1} \), we have:
\[ b_{n-1}^2 = a_{n-1}^2(1 - e_{n-1}^2) \tag{4} \]

### Step 4: Substitute Values
From equations (2) and (3), we can substitute \( b_n \) and \( a_{n-1} \) into equation (4):
\[ b_n^2 = a_n^2(1 - e_n^2) \tag{5} \]
Setting equations (2) and (5) equal gives:
\[ b_{n-1}^2 = a_n^2(1 - e_n^2) \]

### Step 5: Combine Equations
Now, substituting \( a_{n-1} = a_n e_n \) into equation (4):
\[ b_{n-1}^2 = (a_n e_n)^2(1 - e_{n-1}^2) \]
Equating both expressions for \( b_{n-1}^2 \):
\[ a_n^2(1 - e_n^2) = a_n^2 e_n^2(1 - e_{n-1}^2) \]
Dividing both sides by \( a_n^2 \) (assuming \( a_n \neq 0 \)):
\[ 1 - e_n^2 = e_n^2(1 - e_{n-1}^2) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 1 - e_n^2 = e_n^2 - e_n^2 e_{n-1}^2 \]
This leads to:
\[ e_n^2(1 + e_{n-1}^2) = 1 \]
Thus, we can express \( e_n^2 \) in terms of \( e_{n-1}^2 \):
\[ e_n^2 = \frac{1}{1 + e_{n-1}^2} \tag{6} \]

### Step 7: Locus of \( (e_n^2, e_{n-1}^2) \)
Let \( x = e_n^2 \) and \( y = e_{n-1}^2 \). From equation (6):
\[ x = \frac{1}{1 + y} \]
Rearranging gives:
\[ x(1 + y) = 1 \implies xy + x - 1 = 0 \]

### Step 8: Identify the Type of Curve
The equation \( xy + x - 1 = 0 \) can be rewritten as:
\[ xy + x + 0y - 1 = 0 \]
This is a hyperbola in the \( xy \)-plane.

### Conclusion
Thus, the locus of \( (e_n^2, e_{n-1}^2) \) is a rectangular hyperbola.