To solve the problem step by step, we will find the area of triangle \( \Delta ALL' \) where \( L(1, 3) \) and \( L'(1, -1) \) are the ends of the latus rectum of a parabola and \( A \) is the vertex of the parabola.
### Step 1: Find the coordinates of the focus
The coordinates of the points \( L \) and \( L' \) are given as:
- \( L(1, 3) \)
- \( L'(1, -1) \)
The focus \( F \) of the parabola is the midpoint of the line segment joining \( L \) and \( L' \). We can find the midpoint using the formula:
\[
F\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Substituting the coordinates of \( L \) and \( L' \):
\[
F\left( \frac{1 + 1}{2}, \frac{3 + (-1)}{2} \right) = F\left( 1, \frac{2}{2} \right) = F(1, 1)
\]
### Step 2: Determine the length of the latus rectum
The length of the latus rectum is the distance between points \( L \) and \( L' \). We can calculate this using the distance formula:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the coordinates of \( L \) and \( L' \):
\[
\text{Distance} = \sqrt{(1 - 1)^2 + (3 - (-1))^2} = \sqrt{0 + (3 + 1)^2} = \sqrt{4^2} = 4
\]
The length of the latus rectum is \( 4 \), which is equal to \( 4a \). Thus, we can find \( a \):
\[
4a = 4 \implies a = 1
\]
### Step 3: Find the coordinates of the vertex
The vertex \( A \) of the parabola lies on the axis of symmetry, which is vertical in this case. The y-coordinate of the vertex is the same as the y-coordinate of the focus, which is \( 1 \). The distance \( a \) from the focus to the vertex is \( 1 \), so the x-coordinate of the vertex is:
\[
x = 1 - a = 1 - 1 = 0
\]
Thus, the coordinates of the vertex \( A \) are \( A(0, 1) \).
### Step 4: Calculate the area of triangle \( \Delta ALL' \)
The area of triangle \( \Delta ALL' \) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is the length of the latus rectum, which is \( 4 \), and the height is the distance from the vertex \( A \) to the line segment \( LL' \) (which is the y-coordinate of the vertex minus the y-coordinate of the focus):
\[
\text{Height} = |y_A - y_F| = |1 - 1| = 0
\]
However, since we are looking for the area of triangle \( \Delta ALL' \), we can also use the coordinates of the points:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting \( A(0, 1) \), \( L(1, 3) \), and \( L'(1, -1) \):
\[
\text{Area} = \frac{1}{2} \left| 0(3 - (-1)) + 1(-1 - 1) + 1(1 - 3) \right|
\]
\[
= \frac{1}{2} \left| 0 + 1(-2) + 1(-2) \right| = \frac{1}{2} \left| -2 - 2 \right| = \frac{1}{2} \times 4 = 2
\]
### Final Answer
The area of triangle \( \Delta ALL' \) is \( 2 \) square units.
