The vertices of the triangle ABC are `A(0, 0), B(3, 0) and C(3, 4)`, where A and C are foci of an ellipse and B lies on the ellipse. If the length of the latus rectum of the ellipse is `(12)/(p)` units, then the vlaue of p is

Answer

To solve the problem, we need to find the value of \( p \) given the vertices of triangle \( ABC \) with points \( A(0, 0) \), \( B(3, 0) \), and \( C(3, 4) \). The points \( A \) and \( C \) are the foci of an ellipse, and point \( B \) lies on the ellipse. We also know that the length of the latus rectum of the ellipse is given as \( \frac{12}{p} \).

### Step-by-Step Solution:

1. **Identify the foci and the point on the ellipse:**
- The foci of the ellipse are \( A(0, 0) \) and \( C(3, 4) \).
- The point \( B(3, 0) \) lies on the ellipse.

2. **Calculate the distances from point \( B \) to the foci \( A \) and \( C \):**
- The distance \( BA \) (from \( B \) to \( A \)):
\[ BA = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{3^2} = 3 \]
- The distance \( BC \) (from \( B \) to \( C \)):
\[ BC = \sqrt{(3 - 3)^2 + (0 - 4)^2} = \sqrt{(-4)^2} = 4 \]

3. **Using the property of the ellipse:**
- The sum of the distances from any point on the ellipse to the foci is constant and equal to the length of the major axis \( 2a \):
\[ BA + BC = 2a \]
- Substituting the values we found:
\[ 3 + 4 = 2a \implies 2a = 7 \implies a = \frac{7}{2} \]

4. **Relate the length of the latus rectum to \( a \) and \( b \):**
- The length of the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
- According to the problem, the length of the latus rectum is \( \frac{12}{p} \):
\[ \frac{12}{p} = \frac{2b^2}{a} \]

5. **Substituting \( a \) into the equation:**
- We know \( a = \frac{7}{2} \), so substituting this into the equation gives:
\[ \frac{12}{p} = \frac{2b^2}{\frac{7}{2}} \implies \frac{12}{p} = \frac{4b^2}{7} \]

6. **Rearranging to find \( p \):**
- Cross-multiplying gives:
\[ 12 \cdot 7 = 4b^2 \cdot p \implies 84 = 4b^2 p \implies p = \frac{84}{4b^2} = \frac{21}{b^2} \]

7. **Finding \( b^2 \):**
- To find \( b^2 \), we need to use the relationship between \( a \), \( b \), and \( c \) (where \( c \) is the distance from the center to the foci):
\[ c = \sqrt{a^2 - b^2} \]
- The distance between the foci \( A \) and \( C \) is \( 3 \) (the distance between \( (0,0) \) and \( (3,4) \)):
\[ c = \frac{3}{2} \implies c = \sqrt{a^2 - b^2} \implies \left(\frac{3}{2}\right)^2 = \left(\frac{7}{2}\right)^2 - b^2 \]
- Solving gives:
\[ \frac{9}{4} = \frac{49}{4} - b^2 \implies b^2 = \frac{49}{4} - \frac{9}{4} = \frac{40}{4} = 10 \]

8. **Substituting \( b^2 \) back to find \( p \):**
- Now substituting \( b^2 = 10 \) into the equation for \( p \):
\[ p = \frac{21}{10} \]

### Final Answer:
The value of \( p \) is \( \frac{21}{10} \).