If `e_(1)` is the eccentricity of the hyperbola `(x^(2))/(36)-(y^2)/(49) =1` and `e_(2)` is the eccentricity of the hyperbola `(x^(2))/(36)-(y^2)/(49) =-1` then

Options

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

Answer

To solve the problem, we need to find the eccentricities \( e_1 \) and \( e_2 \) of the given hyperbolas and then analyze their relationship.

### Step 1: Identify the hyperbolas
The first hyperbola is given by:
\[ \frac{x^2}{36} - \frac{y^2}{49} = 1 \]
The second hyperbola is given by:
\[ \frac{x^2}{36} - \frac{y^2}{49} = -1 \]

### Step 2: Determine the values of \( a^2 \) and \( b^2 \)
From the first hyperbola, we can identify:
- \( a^2 = 36 \) (thus \( a = 6 \))
- \( b^2 = 49 \) (thus \( b = 7 \))

### Step 3: Calculate the eccentricity \( e_1 \) for the first hyperbola
The formula for the eccentricity of a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:
\[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e_1 = \sqrt{1 + \frac{49}{36}} = \sqrt{1 + \frac{49}{36}} = \sqrt{\frac{36 + 49}{36}} = \sqrt{\frac{85}{36}} = \frac{\sqrt{85}}{6} \]

### Step 4: Calculate the eccentricity \( e_2 \) for the second hyperbola
The formula for the eccentricity of a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \) is:
\[ e_2 = \sqrt{1 + \frac{a^2}{b^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e_2 = \sqrt{1 + \frac{36}{49}} = \sqrt{1 + \frac{36}{49}} = \sqrt{\frac{49 + 36}{49}} = \sqrt{\frac{85}{49}} = \frac{\sqrt{85}}{7} \]

### Step 5: Analyze the relationship between \( e_1 \) and \( e_2 \)
Now we have:
- \( e_1 = \frac{\sqrt{85}}{6} \)
- \( e_2 = \frac{\sqrt{85}}{7} \)

To find the relationship between \( e_1 \) and \( e_2 \), we can calculate \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \):
\[ \frac{1}{e_1^2} = \frac{36}{85}, \quad \frac{1}{e_2^2} = \frac{49}{85} \]
Adding these:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{36}{85} + \frac{49}{85} = \frac{85}{85} = 1 \]

### Conclusion
Thus, we find that:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \]