Let `F_1,F_2` are the foci of the hyperbola `x^2/16-y^2/9=1` and `F_3,F_4` are the foci of its conjugate hyperbola. If `e_H` and `e_C` are their eccentricities respectivley then the statement which holds true is:

Options

A. `their equations of their asymptots are different
B. `e_(h)gte_(c)`
C. `area of the quadrilateral formed by their foci is 50sq. Units
D. their auciliary circles will have the same equation

Answer

To solve the problem, we need to find the eccentricities of the hyperbola and its conjugate hyperbola given by the equation \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \).

### Step 1: Identify the parameters of the hyperbola

The standard form of a hyperbola is given by:

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

From the equation \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \), we can identify:

- \( a^2 = 16 \) ⇒ \( a = 4 \)
- \( b^2 = 9 \) ⇒ \( b = 3 \)

### Step 2: Calculate the eccentricity of the hyperbola

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

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

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

\[ e_H = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{16 + 9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 3: Identify the parameters of the conjugate hyperbola

The conjugate hyperbola has the equation:

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

For our hyperbola, the conjugate hyperbola will be:

\[ \frac{y^2}{9} - \frac{x^2}{16} = 1 \]

### Step 4: Calculate the eccentricity of the conjugate hyperbola

The eccentricity \( e_C \) of the conjugate hyperbola is given by:

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

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

\[ e_C = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9 + 16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]

### Step 5: Compare the eccentricities

Now we have:

- \( e_H = \frac{5}{4} \)
- \( e_C = \frac{5}{3} \)

### Conclusion

The eccentricity of the hyperbola \( e_H \) is less than the eccentricity of the conjugate hyperbola \( e_C \):

\[ e_H < e_C \quad \text{(or)} \quad \frac{5}{4} < \frac{5}{3} \]

Thus, the statement that holds true is that the eccentricity of the hyperbola is less than the eccentricity of its conjugate hyperbola.