To find the length of the latus rectum \( L \) of the ellipse given by the equation
\[
\frac{x^2}{A^2} + \frac{y^2}{9} = 1
\]
where \( A^2 \) is the area enclosed by the quadrilateral formed by joining the foci of the hyperbola
\[
\frac{x^2}{16} - \frac{y^2}{9} = 1
\]
and its conjugate hyperbola, follow these steps:
### Step 1: Identify parameters of the hyperbola
The hyperbola is given by
\[
\frac{x^2}{16} - \frac{y^2}{9} = 1
\]
From this, we can identify \( a^2 = 16 \) and \( b^2 = 9 \).
### Step 2: Calculate the eccentricity of the hyperbola
The eccentricity \( e \) of the hyperbola is given by the formula:
\[
e = \sqrt{1 + \frac{b^2}{a^2}}
\]
Substituting the values of \( a^2 \) and \( b^2 \):
\[
e = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{16 + 9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}
\]
### Step 3: Determine the foci of the hyperbola
The foci of the hyperbola are located at \( (ae, 0) \) and \( (-ae, 0) \). Here, \( a = \sqrt{16} = 4 \), so:
\[
foci = (4 \cdot \frac{5}{4}, 0) = (5, 0) \quad \text{and} \quad (-5, 0)
\]
### Step 4: Determine the foci of the conjugate hyperbola
The foci of the conjugate hyperbola are located at \( (0, be) \) and \( (0, -be) \). Here, \( b = \sqrt{9} = 3 \):
\[
foci = (0, 3 \cdot \frac{5}{4}) = (0, \frac{15}{4}) \quad \text{and} \quad (0, -\frac{15}{4})
\]
### Step 5: Form the quadrilateral and calculate its area
The quadrilateral formed by these four points is a rectangle with vertices at \( (5, 0) \), \( (-5, 0) \), \( (0, \frac{15}{4}) \), and \( (0, -\frac{15}{4}) \).
The area \( A \) of this rectangle is given by:
\[
A = \text{length} \times \text{width} = (5 - (-5)) \times \left(\frac{15}{4} - (-\frac{15}{4})\right) = 10 \times \frac{30}{4} = 10 \times 7.5 = 75
\]
### Step 6: Set \( A^2 \) equal to the area calculated
Since \( A^2 \) is the area of the quadrilateral, we have:
\[
A^2 = 75 \implies A = \sqrt{75} = 5\sqrt{3}
\]
### Step 7: Substitute \( A^2 \) into the ellipse equation
Now, we substitute \( A^2 = 75 \) into the ellipse equation:
\[
\frac{x^2}{75} + \frac{y^2}{9} = 1
\]
### Step 8: Calculate the length of the latus rectum \( L \)
The formula for the length of the latus rectum \( L \) of an ellipse is:
\[
L = \frac{2b^2}{a}
\]
Here, \( b^2 = 9 \) and \( a = \sqrt{75} = 5\sqrt{3} \):
\[
L = \frac{2 \cdot 9}{5\sqrt{3}} = \frac{18}{5\sqrt{3}}
\]
### Step 9: Rationalize the denominator
To rationalize the denominator:
\[
L = \frac{18\sqrt{3}}{15} = \frac{6\sqrt{3}}{5}
\]
Thus, the length of the latus rectum \( L \) is:
\[
L = \frac{18}{5\sqrt{3}}
\]
