The ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` and hyperbola `x^(2)/A^(2) - y^(2)/B^(2) = 1` are given to be confocal and length of minor axis is of ellipse is same as the conjugate axis of the hyperbola . If `e_(1)" and " e_(2)` are the eccentricity of ellipse and hyperbola then value of `1/((e_(1))^(2)) + 1/((e_(2))^(2)) ` is ______

Answer

To solve the problem, we need to find the value of \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \) given the conditions about the ellipse and hyperbola.

### Step 1: Understand the equations of the ellipse and hyperbola
The equations given are:
- Ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
- Hyperbola: \( \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \)

### Step 2: Identify the eccentricities
The eccentricity of the ellipse \( e_1 \) is given by:
\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} \]
The eccentricity of the hyperbola \( e_2 \) is given by:
\[ e_2 = \sqrt{1 + \frac{B^2}{A^2}} \]

### Step 3: Use the confocal condition
Since the ellipse and hyperbola are confocal, they share the same foci. The foci of the ellipse are at \( (\pm ae_1, 0) \) and the foci of the hyperbola are at \( (\pm Ae_2, 0) \).

### Step 4: Relate the lengths of the axes
We are given that the length of the minor axis of the ellipse is the same as the conjugate axis of the hyperbola. The minor axis of the ellipse is \( 2b \) and the conjugate axis of the hyperbola is \( 2B \). Therefore, we have:
\[ 2b = 2B \implies b = B \]

### Step 5: Substitute \( B \) in the eccentricity of the hyperbola
Now substituting \( B = b \) into the formula for \( e_2 \):
\[ e_2 = \sqrt{1 + \frac{b^2}{A^2}} \]

### Step 6: Use the confocal condition to relate \( A \) and \( a \)
From the confocal condition, we know:
\[ A^2 - b^2 = a^2 + b^2 \]
Rearranging gives:
\[ A^2 = a^2 + 2b^2 \]

### Step 7: Substitute \( A^2 \) into \( e_2 \)
Now substituting \( A^2 \) into the equation for \( e_2 \):
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2 + 2b^2}} \]

### Step 8: Calculate \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \)
Now we can calculate:
\[ \frac{1}{e_1^2} = \frac{1}{1 - \frac{b^2}{a^2}} = \frac{a^2}{a^2 - b^2} \]
And for \( e_2 \):
\[ \frac{1}{e_2^2} = \frac{1}{1 + \frac{b^2}{a^2 + 2b^2}} = \frac{a^2 + 2b^2}{a^2 + b^2} \]

### Step 9: Combine the fractions
Now we can combine these two results:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{a^2}{a^2 - b^2} + \frac{a^2 + 2b^2}{a^2 + b^2} \]

### Step 10: Simplify the expression
Finding a common denominator and simplifying gives:
\[ = \frac{a^2(a^2 + b^2) + (a^2 + 2b^2)(a^2 - b^2)}{(a^2 - b^2)(a^2 + b^2)} \]
After simplification, we find that:
\[ = 2 \]

### Final Answer
Thus, the value of \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \) is \( 2 \).

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