The locus of a point which moves such that the difference of its distances from the points (5, 0) and `(-5, 0)` is 6 units is a conic, whose length of the latus rectum (in units) is equal to

Options

A. 4
B. `(16)/(3)`
C. 8
D. `(32)/(3)`

Answer

To solve the problem, we need to find the length of the latus rectum of the hyperbola defined by the locus of points such that the difference of their distances from the points (5, 0) and (-5, 0) is 6 units.

### Step-by-Step Solution:

1. **Identify the foci of the hyperbola**:
The foci of the hyperbola are given as \( F_1(5, 0) \) and \( F_2(-5, 0) \).

2. **Use the definition of hyperbola**:
The definition of a hyperbola states that the difference of the distances from any point \( P(x, y) \) on the hyperbola to the two foci is constant. In this case, the constant difference is 6:
\[ |PF_1 - PF_2| = 6 \]

3. **Set up the equation**:
Let \( PF_1 = d_1 \) and \( PF_2 = d_2 \). We have:
\[ d_1 - d_2 = 6 \]
This implies:
\[ d_1 = d_2 + 6 \]

4. **Distance formula**:
Using the distance formula, we can express \( d_1 \) and \( d_2 \):
\[ d_1 = \sqrt{(x - 5)^2 + y^2} \]
\[ d_2 = \sqrt{(x + 5)^2 + y^2} \]

5. **Substitute into the equation**:
Substitute \( d_1 \) and \( d_2 \) into the equation:
\[ \sqrt{(x - 5)^2 + y^2} - \sqrt{(x + 5)^2 + y^2} = 6 \]

6. **Square both sides to eliminate the square roots**:
Rearranging gives:
\[ \sqrt{(x - 5)^2 + y^2} = \sqrt{(x + 5)^2 + y^2} + 6 \]
Squaring both sides:
\[ (x - 5)^2 + y^2 = ((x + 5)^2 + y^2) + 12\sqrt{(x + 5)^2 + y^2} + 36 \]

7. **Simplify the equation**:
Cancel \( y^2 \) from both sides:
\[ (x - 5)^2 = (x + 5)^2 + 12\sqrt{(x + 5)^2 + y^2} + 36 \]
Expanding and simplifying leads to:
\[ -20x = 12\sqrt{(x + 5)^2 + y^2} + 36 \]

8. **Rearranging gives us the hyperbola equation**:
After further manipulation, we can derive the standard form of the hyperbola:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]
Here, \( a^2 = 9 \) and \( b^2 = 16 \).

9. **Find the length of the latus rectum**:
The length of the latus rectum \( L \) of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting \( a = 3 \) and \( b = 4 \):
\[ L = \frac{2 \cdot 16}{3} = \frac{32}{3} \]

### Final Answer:
The length of the latus rectum of the hyperbola is \( \frac{32}{3} \) units.