Integral part of the area of figure bounded by the tangents at the end of latus rectum of ellipse `(x^(2))/9+(y^(2))/4=1` and directices of hyperbola `(x^(2))/9+(y^(2))/72=1`is

Explanation

To find the integral part of the area of the figure bounded by the tangents at the ends of the latus rectum of the ellipse and the directrices of the hyperbola, we will follow these steps:

### Step 1: Identify the ellipse and hyperbola equations
The equations given are:
- Ellipse: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\)
- Hyperbola: \(\frac{x^2}{9} - \frac{y^2}{72} = 1\)

### Step 2: Find the latus rectum of the ellipse
For the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 9\) and \(b^2 = 4\):
- \(a = 3\)
- \(b = 2\)

The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3} \]

### Step 3: Find the coordinates of the ends of the latus rectum
The ends of the latus rectum are located at:
\[ \left(\pm \frac{b^2}{a}, \pm b\right) = \left(\pm \frac{4}{3}, \pm 2\right) \]
Thus, the points are \(\left(\frac{4}{3}, 2\right)\) and \(\left(-\frac{4}{3}, 2\right)\).

### Step 4: Find the equations of the tangents at these points
The equation of the tangent to the ellipse at point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{9} + \frac{yy_1}{4} = 1 \]
For the point \(\left(\frac{4}{3}, 2\right)\):
\[ \frac{4x}{27} + \frac{2y}{4} = 1 \implies 4x + 27y = 27 \]
For the point \(\left(-\frac{4}{3}, 2\right)\):
\[ \frac{-4x}{27} + \frac{2y}{4} = 1 \implies -4x + 27y = 27 \]

### Step 5: Find the directrices of the hyperbola
The directrices of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by:
\[ x = \pm \frac{a^2}{c} \quad \text{where } c = \sqrt{a^2 + b^2} \]
Here, \(c = \sqrt{9 + 72} = \sqrt{81} = 9\).
Thus, the directrices are:
\[ x = \pm \frac{9}{9} = \pm 1 \]

### Step 6: Find the area of the bounded region
The area of the figure bounded by the tangents and the directrices can be calculated using the coordinates of the intersection points of the tangents and directrices.

1. **Intersection with \(x = 1\)**:
- For \(4x + 27y = 27\):
\[ 4(1) + 27y = 27 \implies 27y = 23 \implies y = \frac{23}{27} \]
- For \(-4x + 27y = 27\):
\[ -4(1) + 27y = 27 \implies 27y = 31 \implies y = \frac{31}{27} \]

2. **Intersection with \(x = -1\)**:
- For \(4(-1) + 27y = 27\):
\[ -4 + 27y = 27 \implies 27y = 31 \implies y = \frac{31}{27} \]
- For \(-4(-1) + 27y = 27\):
\[ 4 + 27y = 27 \implies 27y = 23 \implies y = \frac{23}{27} \]

### Step 7: Calculate the area of the bounded figure
The area can be calculated using the formula for the area of a trapezoid:
\[ \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h \]
Where \(b_1\) and \(b_2\) are the lengths of the bases (heights at \(x = 1\) and \(x = -1\)), and \(h\) is the distance between the bases.

### Step 8: Final calculation and integral part
After calculating the area, we find the integral part of the area.