Let `A=(-3, 4) and B=(2, -1)` be two fixed points. A point C moves such that `tan((1)/(2)angleABC):tan((1)/(2)angleBAC)=3:1`
Thus, locus of C is a hyperbola, distance between whose foci is

Options

A. `5`
B. `5sqrt(2)`
C. `(5)/(2)`
D. `(5)/(sqrt(2))`

Answer

To solve the problem, we need to find the distance between the foci of the hyperbola defined by the given conditions. Let's break down the solution step by step.

### Step 1: Understand the Given Points
We have two fixed points:
- \( A = (-3, 4) \)
- \( B = (2, -1) \)

Let the coordinates of point \( C \) be \( (h, k) \).

### Step 2: Use the Given Ratio of Tangents
We are given that:
\[ \tan\left(\frac{1}{2} \angle ABC\right) : \tan\left(\frac{1}{2} \angle BAC\right) = 3 : 1 \]

Let:
- \( \tan\left(\frac{1}{2} \angle ABC\right) = 3x \)
- \( \tan\left(\frac{1}{2} \angle BAC\right) = x \)

### Step 3: Apply the Formula for Tangents
Using the formula for the tangent of half angles in a triangle, we have:
\[ \tan\left(\frac{1}{2} \angle ABC\right) = \frac{\Delta}{s(s - b)} \]
\[ \tan\left(\frac{1}{2} \angle BAC\right) = \frac{\Delta}{s(s - a)} \]
where \( \Delta \) is the area of triangle \( ABC \), and \( s \) is the semi-perimeter given by:
\[ s = \frac{a + b + c}{2} \]

### Step 4: Set Up the Equation
From the ratio:
\[ \frac{\tan\left(\frac{1}{2} \angle ABC\right)}{\tan\left(\frac{1}{2} \angle BAC\right)} = \frac{3x}{x} = 3 \]
This leads to:
\[ \frac{\Delta / (s(s - b))}{\Delta / (s(s - a))} = 3 \]
Cancelling \( \Delta \) and \( s \):
\[ \frac{s - a}{s - b} = 3 \]

### Step 5: Solve for \( s \)
Cross-multiplying gives:
\[ s - a = 3(s - b) \]
Expanding and rearranging:
\[ s - a = 3s - 3b \implies 2s = 3b + a \implies s = \frac{3b + a}{2} \]

### Step 6: Substitute for \( s \)
Substituting \( s \) back into the semi-perimeter formula:
\[ s = \frac{a + b + c}{2} \implies \frac{3b + a}{2} = \frac{a + b + c}{2} \]
This simplifies to:
\[ 3b + a = a + b + c \implies 2b = c \]

### Step 7: Find the Distance Between Foci
The distance between the foci of a hyperbola is given by \( 2c \), where \( c \) is the distance from the center to each focus. We need to find the distance \( AB \):
\[ AB = \sqrt{(2 - (-3))^2 + (-1 - 4)^2} = \sqrt{(2 + 3)^2 + (-1 - 4)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 8: Conclusion
The distance between the foci of the hyperbola is:
\[ \text{Distance between foci} = 5\sqrt{2} \]