If `e_(1)ande_(2)` respectively be the eccentricities of the ellipse `(x^(2))/(25)+(y^(2))/(9)=1` and hyperbola `9x^(2)-16y^(2)=144` then `e_(1)e_(2)` =

Options

A. `16//25`
B. 1
C. `gt1`
D. `lt1//2`

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: Identify the equations of the ellipse and hyperbola
The equations given are:
1. Ellipse: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)
2. Hyperbola: \(9x^2 - 16y^2 = 144\)

### Step 2: Rewrite the hyperbola in standard form
To rewrite the hyperbola in standard form, we divide the entire equation by 144:
\[ \frac{9x^2}{144} - \frac{16y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

### Step 3: Identify \(a^2\) and \(b^2\) for both conics
For the ellipse:
- \(a^2 = 25\) (major axis)
- \(b^2 = 9\)

For the hyperbola:
- \(a^2 = 16\) (transverse axis)
- \(b^2 = 9\) (conjugate axis)

### Step 4: Calculate the eccentricity of the ellipse (\(e_1\))
The eccentricity \(e_1\) of an ellipse is given by:
\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 5: Calculate the eccentricity of the hyperbola (\(e_2\))
The eccentricity \(e_2\) of a hyperbola is given by:
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{16 + 9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 6: Calculate the product of the eccentricities
Now, we find the product \(e_1 \cdot e_2\):
\[ e_1 \cdot e_2 = \left(\frac{4}{5}\right) \cdot \left(\frac{5}{4}\right) = 1 \]

### Final Answer
Thus, the product of the eccentricities \(e_1 e_2 = 1\).

---