To find the eccentricity of the hyperbola given its foci and that it passes through the origin, we can follow these steps:
### Step 1: Identify the coordinates of the foci
The foci of the hyperbola are given as \( F_1(5, 12) \) and \( F_2(24, 7) \).
### Step 2: Calculate the distance between the foci
The distance \( d \) between the two foci \( F_1 \) and \( F_2 \) can be calculated using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the coordinates:
\[
d = \sqrt{(24 - 5)^2 + (7 - 12)^2} = \sqrt{19^2 + (-5)^2} = \sqrt{361 + 25} = \sqrt{386}
\]
### Step 3: Determine the distance between the foci in terms of \( 2c \)
In a hyperbola, the distance between the foci is given by \( 2c \), where \( c \) is the distance from the center to each focus. Thus:
\[
2c = \sqrt{386} \implies c = \frac{\sqrt{386}}{2}
\]
### Step 4: Use the property of hyperbolas
For a hyperbola that passes through a point \( P(0, 0) \) (the origin), the relationship between the distances from the point to the foci is:
\[
|d(P, F_2) - d(P, F_1)| = 2a
\]
Where \( a \) is the distance from the center to the vertices.
### Step 5: Calculate the distances from the origin to the foci
Calculate \( d(P, F_1) \) and \( d(P, F_2) \):
\[
d(P, F_1) = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]
\[
d(P, F_2) = \sqrt{(24 - 0)^2 + (7 - 0)^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25
\]
### Step 6: Apply the hyperbola property
Now, apply the property:
\[
|25 - 13| = 2a \implies 12 = 2a \implies a = 6
\]
### Step 7: Relate \( a \), \( b \), and \( c \)
In a hyperbola, the relationship between \( a \), \( b \), and \( c \) is given by:
\[
c^2 = a^2 + b^2
\]
We already have \( a = 6 \) and \( c = \frac{\sqrt{386}}{2} \). Thus:
\[
c^2 = \left(\frac{\sqrt{386}}{2}\right)^2 = \frac{386}{4}
\]
\[
a^2 = 6^2 = 36
\]
Now substituting into the equation:
\[
\frac{386}{4} = 36 + b^2 \implies b^2 = \frac{386}{4} - 36 = \frac{386 - 144}{4} = \frac{242}{4} = \frac{121}{2}
\]
### Step 8: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the hyperbola is given by:
\[
e = \frac{c}{a} = \frac{\frac{\sqrt{386}}{2}}{6} = \frac{\sqrt{386}}{12}
\]
### Final Answer
Thus, the eccentricity of the hyperbola is:
\[
e = \frac{\sqrt{386}}{12}
\]
