If `e_(1),e_(2)` are the eccentricites of the hyperbla `2x^(2)-2y^(2)=1` and the ellipse `x^(2)+2y^(2)=2` respectively then

Options

A. `e_(1)+e_(2)=1`
B. `e_(1)e_(2)=1`
C. `e_(1)^(2)+e_(2)^(2)=1`
D. none of these

Answer

To solve the problem, we need to find the eccentricities \( e_1 \) and \( e_2 \) of the given hyperbola and ellipse respectively.

### Step 1: Find the eccentricity of the hyperbola

The equation of the hyperbola is given as:

\[ 2x^2 - 2y^2 = 1 \]

We can rewrite this in standard form by dividing the entire equation by 1:

\[ \frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1 \]

From this, we can identify \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{2} \).

The eccentricity \( e_1 \) of a hyperbola is given by the formula:

\[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting the values of \( a^2 \) and \( b^2 \):

\[ e_1 = \sqrt{1 + \frac{\frac{1}{2}}{\frac{1}{2}}} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 2: Find the eccentricity of the ellipse

The equation of the ellipse is given as:

\[ x^2 + 2y^2 = 2 \]

We can rewrite this in standard form by dividing the entire equation by 2:

\[ \frac{x^2}{2} + \frac{y^2}{1} = 1 \]

From this, we can identify \( a^2 = 2 \) and \( b^2 = 1 \).

The eccentricity \( e_2 \) of an ellipse is given by the formula:

\[ e_2 = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the values of \( a^2 \) and \( b^2 \):

\[ e_2 = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 3: Summarize the results

Now we have:

- \( e_1 = \sqrt{2} \)
- \( e_2 = \frac{1}{\sqrt{2}} \)

### Step 4: Check the relationship between \( e_1 \) and \( e_2 \)

We can check the product of the eccentricities:

\[ e_1 \cdot e_2 = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \]

This shows that the product of the eccentricities is equal to 1.

### Final Answer

Thus, the relationship between the eccentricities \( e_1 \) and \( e_2 \) is:

\[ e_1 \cdot e_2 = 1 \]