Tangents are drawn to the ellipse `x^2/9+y^2/5 = 1` at the end of latus rectum. Find the area of quadrilateral so formed

Options

A. `27/2`
B. 27
C. `27/4`
D. 18

Answer

To solve the problem of finding 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 can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is in the 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 eccentricity of the ellipse
The eccentricity \( e \) of the ellipse can be calculated using the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Step 3: Calculate the coordinates of the ends of the latus rectum
The ends of the latus rectum of an ellipse are given by the points \( (ae, \frac{b^2}{a}) \) and \( (-ae, \frac{b^2}{a}) \). Thus, we find:
- \( ae = 3 \times \frac{2}{3} = 2 \)
- \( \frac{b^2}{a} = \frac{5}{3} \)

The coordinates of the ends of the latus rectum are:
- \( \left(2, \frac{5}{3}\right) \) and \( \left(-2, \frac{5}{3}\right) \)

### Step 4: Write the equations of the tangents at these points
Using the point \( (x_1, y_1) = \left(2, \frac{5}{3}\right) \), the equation of the tangent to the ellipse is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting the values:
\[ \frac{2x}{9} + \frac{3y}{5} = 1 \]
Multiplying through by 45 to eliminate the denominators:
\[ 10x + 27y = 45 \quad \text{(Equation 1)} \]

For the point \( \left(-2, \frac{5}{3}\right) \):
\[ \frac{-2x}{9} + \frac{3y}{5} = 1 \]
Multiplying through by 45:
\[ -10x + 27y = 45 \quad \text{(Equation 2)} \]

### Step 5: Find the intersection points with the axes
**For Equation 1:**
- Set \( x = 0 \):
\[ 27y = 45 \implies y = \frac{45}{27} = \frac{5}{3} \]
- Set \( y = 0 \):
\[ 10x = 45 \implies x = \frac{45}{10} = \frac{9}{2} \]

**For Equation 2:**
- Set \( x = 0 \):
\[ 27y = 45 \implies y = \frac{5}{3} \]
- Set \( y = 0 \):
\[ -10x = 45 \implies x = -\frac{45}{10} = -\frac{9}{2} \]

### Step 6: Determine the vertices of the quadrilateral
The vertices of the quadrilateral formed by the tangents are:
1. \( \left(0, \frac{5}{3}\right) \)
2. \( \left(\frac{9}{2}, 0\right) \)
3. \( \left(0, -\frac{5}{3}\right) \)
4. \( \left(-\frac{9}{2}, 0\right) \)

### Step 7: Calculate the area of the quadrilateral
The area of the quadrilateral can be calculated by finding the area of one triangle formed by the points \( \left(0, \frac{5}{3}\right) \), \( \left(\frac{9}{2}, 0\right) \), and the origin \( (0, 0) \):
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{9}{2} \times \frac{5}{3} = \frac{27}{4} \]
Since the quadrilateral is symmetric, the total area is:
\[ \text{Total Area} = 4 \times \frac{27}{4} = 27 \]

### Final Answer
The area of the quadrilateral formed is \( 27 \) square units.