The locus of a point whose distance from the point (3,0) is 3/5 times its distance from the line x=p is an ellipse with centre at the origin. The value of p is

Options

A. 5
B. 7
C. `25/3`
D. `25/9`

Answer

To solve the problem, we need to find the value of \( p \) such that the locus of a point whose distance from the point \( (3, 0) \) is \( \frac{3}{5} \) times its distance from the line \( x = p \) forms an ellipse centered at the origin.

### Step-by-Step Solution:

1. **Identify the Distances**:
- Let \( P(h, k) \) be any point on the locus.
- The distance from \( P \) to the point \( (3, 0) \) is given by:
\[ d_1 = \sqrt{(h - 3)^2 + k^2} \]
- The distance from \( P \) to the line \( x = p \) is:
\[ d_2 = |h - p| \]

2. **Set Up the Relationship**:
- According to the problem, we have:
\[ d_1 = \frac{3}{5} d_2 \]
- Substituting the distances, we get:
\[ \sqrt{(h - 3)^2 + k^2} = \frac{3}{5} |h - p| \]

3. **Square Both Sides**:
- To eliminate the square root, we square both sides:
\[ (h - 3)^2 + k^2 = \left(\frac{3}{5} |h - p|\right)^2 \]
- This simplifies to:
\[ (h - 3)^2 + k^2 = \frac{9}{25} (h - p)^2 \]

4. **Rearranging the Equation**:
- Multiply through by 25 to clear the fraction:
\[ 25((h - 3)^2 + k^2) = 9(h - p)^2 \]
- Expanding both sides:
\[ 25(h^2 - 6h + 9 + k^2) = 9(h^2 - 2ph + p^2) \]
- This leads to:
\[ 25h^2 - 150h + 225 + 25k^2 = 9h^2 - 18ph + 9p^2 \]

5. **Combine Like Terms**:
- Rearranging gives:
\[ 25h^2 - 9h^2 + 18ph - 150h + 25k^2 + 225 - 9p^2 = 0 \]
- Simplifying further:
\[ 16h^2 + (18p - 150)h + 25k^2 + (225 - 9p^2) = 0 \]

6. **Condition for an Ellipse**:
- For this equation to represent an ellipse, the discriminant must be less than zero. The discriminant \( D \) for the quadratic in \( h \) is:
\[ D = (18p - 150)^2 - 4 \cdot 16 \cdot (25k^2 + 225 - 9p^2) \]
- Setting \( D < 0 \) will give us conditions on \( p \).

7. **Finding the Value of \( p \)**:
- We know that the focus of the ellipse is at \( (3, 0) \) and the eccentricity \( e = \frac{3}{5} \).
- The relationship between the semi-major axis \( a \) and the directrix \( p \) is given by:
\[ p = \frac{a}{e} \]
- From the focus, we have \( a \cdot e = 3 \), thus:
\[ a = \frac{3}{e} = \frac{3}{\frac{3}{5}} = 5 \]
- Therefore:
\[ p = \frac{5}{\frac{3}{5}} = \frac{25}{3} \]

### Final Answer:
The value of \( p \) is \( \frac{25}{3} \).