Consider an ellipse ` x^2/a^2+y^2/b^2=1` Let a hyperbola is having its vertices at the extremities of minor axis of an ellipse and length of major axis of an ellipse is equal to the distance between the foci of hyperbola. Let `e_1` and `e_2` be the eccentricities of an ellipse and hyperbola respectively. Again let A be the area of the quadrilateral formed by joining all the foci and A, be the area of the quadrilateral formed by all the directrices. The relation between `e_1 and e_2` is given by

Options

A. `e_(1)e_(2)=1`
B. `e_(2)^(2)(1-e_(1)^(2))=1`
C. `e_(1)^(2)(e_(1)^(2)-1)=1`
D. `e_(1)e_(2)(1-e_(1)^(2))=1`

Explanation

To solve the given problem, we will derive the relationship between the eccentricities \( e_1 \) (of the ellipse) and \( e_2 \) (of the hyperbola) step by step.

### Step 1: Understanding the Ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The eccentricity \( e_1 \) of the ellipse is defined as:
\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} \]

### Step 2: Foci of the Ellipse
The foci of the ellipse are located at:
\[ (\pm ae_1, 0) = (\pm a\sqrt{1 - \frac{b^2}{a^2}}, 0) \]

### Step 3: Hyperbola Properties
The hyperbola has its vertices at the extremities of the minor axis of the ellipse, which are at:
\[ (0, \pm b) \]
The distance between the foci of the hyperbola is equal to the length of the major axis of the ellipse, which is \( 2a \).

### Step 4: Eccentricity of the Hyperbola
The distance between the foci of the hyperbola is given by:
\[ 2c = 2ae_2 \]
where \( c = ae_2 \). Setting this equal to the length of the major axis of the ellipse, we have:
\[ 2ae_2 = 2a \implies e_2 = 1 \]

### Step 5: Establishing the Relationship
From the eccentricity definitions, we have:
1. For the ellipse:
\[ e_1^2 = 1 - \frac{b^2}{a^2} \]
2. For the hyperbola:
\[ e_2^2 = \frac{a^2}{b^2} + 1 \]

### Step 6: Substituting and Rearranging
Using the relationship \( e_2 = \frac{a}{b} \), we substitute into the equation for \( e_1 \):
\[ e_2^2 = \frac{a^2}{b^2} = \frac{1 - e_1^2}{e_1^2} \]
Rearranging gives us:
\[ e_2^2 - e_1^2 e_2^2 = 1 \]
Factoring out \( e_2^2 \):
\[ e_2^2(1 - e_1^2) = 1 \]

### Conclusion
Thus, the relationship between the eccentricities \( e_1 \) and \( e_2 \) is given by:
\[ e_2^2(1 - e_1^2) = 1 \]