To find the area of the quadrilateral formed by the tangents drawn to the ellipse \(\frac{x^2}{9} + \frac{y^2}{5} = 1\) at the ends of the latus rectum, we will follow these steps:
### Step 1: Identify the parameters of the ellipse
The given ellipse is in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 9\) and \(b^2 = 5\). Thus, we have:
- \(a = 3\)
- \(b = \sqrt{5}\)
### Step 2: Find the coordinates of the ends of the latus rectum
The ends of the latus rectum for an ellipse are given by the points \((\pm ae, \frac{b^2}{a})\), where \(e\) is the eccentricity.
First, we calculate the eccentricity \(e\):
\[
e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}
\]
Now, we can find the coordinates of the ends of the latus rectum:
\[
\text{Coordinates} = \left(\pm ae, \frac{b^2}{a}\right) = \left(\pm 3 \cdot \frac{2}{3}, \frac{5}{3}\right) = \left(\pm 2, \frac{5}{3}\right)
\]
Thus, the points are \((2, \frac{5}{3})\) and \((-2, \frac{5}{3})\).
### Step 3: Find the equations of the tangents at these points
The equation of the tangent to the ellipse at the point \((x_1, y_1)\) is given by:
\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1
\]
For the point \((2, \frac{5}{3})\):
\[
\frac{2x}{9} + \frac{y \cdot \frac{5}{3}}{5} = 1
\]
Simplifying this, we get:
\[
\frac{2x}{9} + \frac{y}{3} = 1 \quad \Rightarrow \quad 2x + 3y = 9 \quad \text{(Equation 1)}
\]
For the point \((-2, \frac{5}{3})\):
\[
\frac{-2x}{9} + \frac{y \cdot \frac{5}{3}}{5} = 1
\]
Simplifying this, we get:
\[
-\frac{2x}{9} + \frac{y}{3} = 1 \quad \Rightarrow \quad -2x + 3y = 9 \quad \Rightarrow \quad 2x - 3y = -9 \quad \text{(Equation 2)}
\]
### Step 4: Find the points of intersection of the tangents with the axes
**For Equation 1:**
- **X-intercept (set \(y = 0\))**:
\[
2x = 9 \quad \Rightarrow \quad x = \frac{9}{2} \quad \Rightarrow \quad A\left(\frac{9}{2}, 0\right)
\]
- **Y-intercept (set \(x = 0\))**:
\[
3y = 9 \quad \Rightarrow \quad y = 3 \quad \Rightarrow \quad B(0, 3)
\]
**For Equation 2:**
- **X-intercept (set \(y = 0\))**:
\[
2x = -9 \quad \Rightarrow \quad x = -\frac{9}{2} \quad \Rightarrow \quad C\left(-\frac{9}{2}, 0\right)
\]
- **Y-intercept (set \(x = 0\))**:
\[
-3y = -9 \quad \Rightarrow \quad y = 3 \quad \Rightarrow \quad D(0, 3)
\]
### Step 5: Calculate the area of the quadrilateral formed by points A, B, C, and D
The quadrilateral is symmetric about the y-axis, and we can find the area of triangle \(AOB\) and then multiply by 2 for the total area.
Area of triangle \(AOB\):
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}
\]
Thus, the total area of the quadrilateral is:
\[
\text{Total Area} = 2 \times \frac{27}{4} = \frac{27}{2}
\]
### Final Area Calculation
Since we need to consider the area formed by both tangents, we multiply by 2:
\[
\text{Required Area} = 4 \times \frac{27}{4} = 27 \text{ square units}
\]
### Conclusion
The area of the quadrilateral formed by the tangents at the ends of the latus rectum of the ellipse is \(27\) square units.
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