To find the eccentricity of the ellipse given the foci at (5, 12) and (24, 7) and that it passes through the point (0, 0), we can follow these steps:
### Step 1: Identify the foci and calculate the distance between them.
The foci are given as:
- \( S_1 = (5, 12) \)
- \( S_2 = (24, 7) \)
To find the distance between the foci \( S_1 \) and \( S_2 \), we use the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the coordinates of the foci:
\[
d = \sqrt{(24 - 5)^2 + (7 - 12)^2}
\]
\[
= \sqrt{(19)^2 + (-5)^2}
\]
\[
= \sqrt{361 + 25}
\]
\[
= \sqrt{386}
\]
### Step 2: Use the property of the ellipse.
For any point \( P \) on the ellipse, the sum of the distances from \( P \) to the two foci is constant and equal to \( 2a \), where \( a \) is the semi-major axis length. Here, we will calculate the distances from the point \( (0, 0) \) to each focus.
#### Distance from \( (0, 0) \) to \( S_1 \):
\[
PS_1 = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]
#### Distance from \( (0, 0) \) to \( S_2 \):
\[
PS_2 = \sqrt{(24 - 0)^2 + (7 - 0)^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25
\]
### Step 3: Calculate \( 2a \).
Now we can find \( 2a \):
\[
2a = PS_1 + PS_2 = 13 + 25 = 38
\]
### Step 4: Calculate the distance between the foci.
From Step 1, we found that the distance between the foci is \( \sqrt{386} \). The distance between the foci is given by \( 2c \), where \( c \) is the distance from the center to each focus.
\[
2c = \sqrt{386} \implies c = \frac{\sqrt{386}}{2}
\]
### Step 5: Relate \( a \), \( b \), and \( c \).
For an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by:
\[
c^2 = a^2 - b^2
\]
### Step 6: Calculate \( a \) and \( c \).
We already have \( 2a = 38 \), so:
\[
a = 19
\]
Now, we can find \( c \):
\[
c = \frac{\sqrt{386}}{2}
\]
### Step 7: Calculate the eccentricity \( e \).
The eccentricity \( e \) is given by:
\[
e = \frac{c}{a}
\]
Substituting the values we have:
\[
e = \frac{\frac{\sqrt{386}}{2}}{19} = \frac{\sqrt{386}}{38}
\]
### Final Answer:
The eccentricity of the ellipse is:
\[
e = \frac{\sqrt{386}}{38}
\]
