Let S=(3,4) and S'=(9,12) be two foci of an ellipse. If the coordinates of the foot of the perpendicular from focus S to a tangent of the ellipse is (1, -4) then the eccentricity of the ellipse is

Explanation

To find the eccentricity of the ellipse given the foci and the foot of the perpendicular from one focus to the tangent, we can follow these steps:

### Step 1: Identify the coordinates of the foci and the foot of the perpendicular
The foci of the ellipse are given as:
- \( S = (3, 4) \)
- \( S' = (9, 12) \)

The foot of the perpendicular from focus \( S \) to the tangent is given as:
- \( P = (1, -4) \)

### Step 2: Calculate the distance between the two foci
The distance \( 2c \) between the two foci \( S \) and \( S' \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of \( S \) and \( S' \):
\[ 2c = \sqrt{(9 - 3)^2 + (12 - 4)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]
Thus, \( c = 5 \).

### Step 3: Find the center of the ellipse
The center \( O \) of the ellipse is the midpoint of the segment joining the foci \( S \) and \( S' \):
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + 9}{2}, \frac{4 + 12}{2} \right) = \left( 6, 8 \right) \]

### Step 4: Use the foot of the perpendicular to find the semi-major axis \( a \)
The foot of the perpendicular \( P \) lies on the circle with center \( O \) and radius \( a \). We can use the distance from \( O \) to \( P \) to find \( a \):
\[ OP = \sqrt{(1 - 6)^2 + (-4 - 8)^2} = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
Thus, \( a = 13 \).

### Step 5: Relate \( a \), \( b \), and \( c \) to find \( b \)
In an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by:
\[ c^2 = a^2 - b^2 \]
Substituting the values we found:
\[ 5^2 = 13^2 - b^2 \implies 25 = 169 - b^2 \implies b^2 = 169 - 25 = 144 \implies b = 12 \]

### Step 6: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the ellipse is given by:
\[ e = \frac{c}{a} = \frac{5}{13} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{5}{13}} \]