If `e_(1)`, and `e_(2)` are respectively the eccentricities of the conics `(x^(2))/(25)-(y^(2))/(11)=1` and `(x^(2))/(16)+(y^(2))/(7)=1` then `e_(1)e_(2)` is equal to

Options

A. `10/9`
B. `4/3`
C. `9/10`
D. `8/5`

Answer

To solve the problem, we need to find the eccentricities \( e_1 \) and \( e_2 \) of the given conics and then calculate the product \( e_1 e_2 \).

### Step 1: Identify the conics
The first conic is given by the equation:
\[ \frac{x^2}{25} - \frac{y^2}{11} = 1 \]
This is a hyperbola because it has a minus sign between the two terms.

The second conic is given by the equation:
\[ \frac{x^2}{16} + \frac{y^2}{7} = 1 \]
This is an ellipse because both terms are positive.

### Step 2: Calculate the eccentricity \( e_1 \) of the hyperbola
For a hyperbola in the form:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
the eccentricity \( e \) is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Here, \( a^2 = 25 \) and \( b^2 = 11 \).

Calculating \( e_1 \):
\[ e_1 = \sqrt{1 + \frac{11}{25}} = \sqrt{1 + 0.44} = \sqrt{1.44} = \frac{6}{5} \]

### Step 3: Calculate the eccentricity \( e_2 \) of the ellipse
For an ellipse in the form:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
the eccentricity \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Here, \( a^2 = 16 \) and \( b^2 = 7 \).

Calculating \( e_2 \):
\[ e_2 = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{16 - 7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

### Step 4: Calculate the product \( e_1 e_2 \)
Now, we find the product of the two eccentricities:
\[ e_1 e_2 = \left(\frac{6}{5}\right) \left(\frac{3}{4}\right) = \frac{18}{20} = \frac{9}{10} \]

### Final Answer
Thus, the value of \( e_1 e_2 \) is:
\[ \boxed{\frac{9}{10}} \]