The area of the quadrilateral formed by the tangents at the end points of latus rectum of the ellipse `5x^(2) +9y^(2)=45` is

Options

A. 9
B. 27/2
C. 27/4
D. None of these

Explanation

To find the area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse given by the equation \(5x^2 + 9y^2 = 45\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 5x^2 + 9y^2 = 45 \]
Dividing the entire equation by 45, we get:
\[ \frac{x^2}{9} + \frac{y^2}{5} = 1 \]
This shows that \(a^2 = 9\) and \(b^2 = 5\), hence \(a = 3\) and \(b = \sqrt{5}\).

### Step 2: Find the foci of the ellipse
The foci of the ellipse can be found using the formula:
\[ c = \sqrt{a^2 - b^2} \]
Calculating \(c\):
\[ c = \sqrt{9 - 5} = \sqrt{4} = 2 \]
Thus, the foci are located at \((\pm c, 0) = (\pm 2, 0)\).

### Step 3: Determine the endpoints of the latus rectum
The endpoints of the latus rectum of an ellipse are given by the coordinates:
\[ \left( \pm \frac{b^2}{a}, c \right) \quad \text{and} \quad \left( \pm \frac{b^2}{a}, -c \right) \]
Calculating the endpoints:
\[ \frac{b^2}{a} = \frac{5}{3} \]
Thus, the endpoints of the latus rectum are:
\[ \left( \frac{5}{3}, 2 \right), \left( \frac{5}{3}, -2 \right), \left( -\frac{5}{3}, 2 \right), \left( -\frac{5}{3}, -2 \right) \]

### Step 4: Find the equations of the tangents at the endpoints
The equation of the tangent to the ellipse at point \((x_1, y_1)\) is given by:
\[ \frac{b^2 x}{a^2} x_1 + \frac{a^2 y}{b^2} y_1 = 1 \]
Using the points \(\left( \frac{5}{3}, 2 \right)\) and \(\left( \frac{5}{3}, -2 \right)\), we can find the equations of the tangents.

For the point \(\left( \frac{5}{3}, 2 \right)\):
\[ \frac{5}{9} x + \frac{9}{5} y = 1 \]
Multiplying through by 45 to eliminate fractions:
\[ 25x + 81y = 45 \]

For the point \(\left( -\frac{5}{3}, 2 \right)\):
\[ \frac{5}{9} x - \frac{9}{5} y = 1 \]
Multiplying through by 45:
\[ 25x - 81y = 45 \]

### Step 5: Find the area of the quadrilateral
The area of the quadrilateral formed by these tangents can be calculated using the formula for the area of a quadrilateral formed by two pairs of lines:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
In our case, the base is the distance between the two vertical lines (which is \(2 \times \frac{5}{3} = \frac{10}{3}\)) and the height is the distance between the two horizontal lines (which is \(2 + 2 = 4\)).

Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times \frac{10}{3} \times 4 = \frac{20}{3} \]

### Final Area Calculation
The total area of the quadrilateral formed by the tangents at the endpoints of the latus rectum is:
\[ \text{Area} = 2 \times \frac{20}{3} = \frac{40}{3} \]

### Conclusion
The area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse \(5x^2 + 9y^2 = 45\) is \(\frac{40}{3}\) square units.

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