If e and e' are the eccentricities of the ellipse `5x^(2) + 9 y^(2) = 45 ` and the hyperbola `5x^(2) - 4y^(2) = 45 respectively , then ee' is equal to

Options

A. `-1`
B. `1`
C. `-4`
D. `9`

Answer

To solve the problem, we need to find the eccentricities of the given ellipse and hyperbola, and then calculate the product of these eccentricities.

### Step 1: Find the eccentricity of the ellipse
The equation of the ellipse is given as:
\[ 5x^2 + 9y^2 = 45 \]

**Convert to standard form:**
1. Divide the entire equation by 45:
\[ \frac{5x^2}{45} + \frac{9y^2}{45} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + \frac{y^2}{5} = 1 \]

**Identify \(a^2\) and \(b^2\):**
- Here, \(a^2 = 9\) and \(b^2 = 5\).

**Calculate the eccentricity \(e\):**
The formula for the eccentricity of an ellipse is:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Step 2: Find the eccentricity of the hyperbola
The equation of the hyperbola is given as:
\[ 5x^2 - 4y^2 = 45 \]

**Convert to standard form:**
1. Divide the entire equation by 45:
\[ \frac{5x^2}{45} - \frac{4y^2}{45} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{\frac{45}{4}} = 1 \]

**Identify \(a^2\) and \(b^2\):**
- Here, \(a^2 = 9\) and \(b^2 = \frac{45}{4}\).

**Calculate the eccentricity \(e'\):**
The formula for the eccentricity of a hyperbola is:
\[ e' = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values:
\[ e' = \sqrt{1 + \frac{\frac{45}{4}}{9}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Step 3: Calculate the product of the eccentricities
Now we need to find the product \(ee'\):
\[ ee' = \left(\frac{2}{3}\right) \left(\frac{3}{2}\right) = \frac{2 \cdot 3}{3 \cdot 2} = 1 \]

### Final Answer:
The product of the eccentricities \(ee'\) is equal to \(1\).

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