Let `x^2/25+y^2/b^2=1` and `x^2/16-y^2/b^2=1` are ellipse and hyperbola respectivelysuch that `e_1e_2=1` where `e_1` and `e_2` are eccentricities. If distance between foci of ellipse is `alpha` and that of hyperbola is `beta` the `(alpha,beta)=`

Options

A. `(4,5)`
B. `(8,10)`
C. `(10,7)`
D. `(4,10)`

Answer

To solve the problem, we need to find the distances between the foci of the given ellipse and hyperbola, and then express them as \((\alpha, \beta)\).

### Step 1: Identify the parameters of the ellipse and hyperbola

The equations given are:
1. Ellipse: \(\frac{x^2}{25} + \frac{y^2}{b^2} = 1\)
2. Hyperbola: \(\frac{x^2}{16} - \frac{y^2}{b^2} = 1\)

From the ellipse equation, we can identify:
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = b^2\) (we will find \(b\) later)

From the hyperbola equation, we can identify:
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = b^2\) (we will find \(b\) later)

### Step 2: Calculate the eccentricities \(e_1\) and \(e_2\)

For the ellipse, the eccentricity \(e_1\) is given by:
\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{b^2}{25}} \]

For the hyperbola, the eccentricity \(e_2\) is given by:
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{16}} \]

### Step 3: Use the condition \(e_1 e_2 = 1\)

We have the condition:
\[ e_1 e_2 = 1 \]
Substituting the expressions for \(e_1\) and \(e_2\):
\[ \sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1 \]

### Step 4: Square both sides

Squaring both sides gives:
\[ \left(1 - \frac{b^2}{25}\right) \left(1 + \frac{b^2}{16}\right) = 1 \]

Expanding the left-hand side:
\[ 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \]

### Step 5: Simplify the equation

Subtracting 1 from both sides:
\[ \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 0 \]

Finding a common denominator (400):
\[ \frac{25b^2 - 16b^2 - b^4}{400} = 0 \]
\[ (9b^2 - b^4) = 0 \]

Factoring out \(b^2\):
\[ b^2(9 - b^2) = 0 \]

This gives us:
\[ b^2 = 0 \quad \text{or} \quad b^2 = 9 \]

Since \(b^2\) cannot be zero, we have:
\[ b^2 = 9 \Rightarrow b = 3 \]

### Step 6: Calculate eccentricities

Now substituting \(b^2 = 9\) back into the eccentricity formulas:
\[ e_1 = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
\[ e_2 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 7: Calculate distances between foci

The distance between the foci of the ellipse is given by:
\[ \alpha = 2a e_1 = 2 \cdot 5 \cdot \frac{4}{5} = 8 \]

The distance between the foci of the hyperbola is given by:
\[ \beta = 2a e_2 = 2 \cdot 4 \cdot \frac{5}{4} = 10 \]

### Final Result

Thus, the final answer is:
\[ (\alpha, \beta) = (8, 10) \]