If foci of `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` coincide with the foci of `(x^(2))/(25)+(y^(2))/(16)=1` and eccentricity of the hyperbola is 3. then

Options

A. `a^(2)+b^(2)=9`
B. there is no directrix circle to the hyperbola
C. centre of the directrix circle is (0, 0)
D. Length of the latusrecum of the hyperbola =16

Answer

To solve the problem, we need to analyze the given hyperbola and the ellipse, and find the required parameters step by step.

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse is given as:

\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

From this, we can identify:
- \( a^2 = 25 \) which gives \( a = 5 \)
- \( b^2 = 16 \) which gives \( b = 4 \)

### Step 2: Calculate the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse can be calculated using the formula:

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

Substituting the values:

\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Determine the foci of the ellipse
The foci of the ellipse are given by:

\[ (\pm ae, 0) = (\pm 5 \cdot \frac{3}{5}, 0) = (\pm 3, 0) \]

### Step 4: Set up the hyperbola
The equation of the hyperbola is given as:

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

The foci of the hyperbola are also given by:

\[ (\pm ae, 0) \]

### Step 5: Use the given eccentricity of the hyperbola
The eccentricity of the hyperbola is given as \( e = 3 \). Therefore, we have:

\[ ae = 3 \]

### Step 6: Relate \( a \) and \( e \)
From the eccentricity relation, we can express \( a \):

\[ a \cdot 3 = 3 \implies a = 1 \]

### Step 7: Find \( b^2 \) using the eccentricity formula
The eccentricity of the hyperbola is also given by:

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

Substituting \( e = 3 \) and \( a = 1 \):

\[ 3^2 = 1 + \frac{b^2}{1^2} \implies 9 = 1 + b^2 \implies b^2 = 8 \]

### Step 8: Check the options
1. **Check if \( a^2 + b^2 = 9 \)**:
\[ a^2 + b^2 = 1 + 8 = 9 \quad \text{(True)} \]

2. **Check for the existence of the director circle**:
The director circle exists if \( a^2 > b^2 \). Here, \( 1 < 8 \) (False).

3. **Center of the director circle**:
Since there is no director circle, this option is also False.

4. **Length of the latus rectum**:
The length of the latus rectum of the hyperbola is given by \( \frac{2b^2}{a} \):
\[ \text{Length} = \frac{2 \cdot 8}{1} = 16 \quad \text{(True)} \]

### Final Results
- Option 1: True
- Option 2: False
- Option 3: False
- Option 4: True