A set of points is such that each point is three times as far away from the y-axis as it is from the point (4,0).Then locus of the points is:

Options

A. hyperbola
B. parabola
C. ellipse
D. circle

Answer

To solve the problem step by step, we need to find the locus of points such that each point is three times as far away from the y-axis as it is from the point (4, 0).

### Step-by-Step Solution:

1. **Define the Points**:
Let \( P(x, y) \) be any point in the plane. The distance from this point to the y-axis is given by the absolute value of its x-coordinate, which is \( |x| \). The distance from point \( P \) to the point \( (4, 0) \) is calculated using the distance formula:
\[ PS = \sqrt{(x - 4)^2 + (y - 0)^2} = \sqrt{(x - 4)^2 + y^2} \]

2. **Set Up the Relationship**:
According to the problem, the distance from point \( P \) to the y-axis is three times the distance from point \( P \) to the point \( (4, 0) \):
\[ |x| = 3 \cdot PS \]
Substituting the expression for \( PS \):
\[ |x| = 3 \sqrt{(x - 4)^2 + y^2} \]

3. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ x^2 = 9 \left( (x - 4)^2 + y^2 \right) \]

4. **Expand and Rearrange**:
Expanding the right side:
\[ x^2 = 9 \left( x^2 - 8x + 16 + y^2 \right) \]
This simplifies to:
\[ x^2 = 9x^2 - 72x + 144 + 9y^2 \]
Rearranging gives:
\[ 0 = 8x^2 - 72x + 144 + 9y^2 \]

5. **Divide by 8**:
To simplify, divide the entire equation by 8:
\[ 0 = x^2 - 9x + 18 + \frac{9}{8}y^2 \]

6. **Complete the Square**:
Completing the square for the \( x \) terms:
\[ 0 = \left( x - \frac{9}{2} \right)^2 - \frac{81}{4} + 18 + \frac{9}{8}y^2 \]
Simplifying gives:
\[ 0 = \left( x - \frac{9}{2} \right)^2 + \frac{9}{8}y^2 - \frac{81}{4} + \frac{72}{4} \]
\[ 0 = \left( x - \frac{9}{2} \right)^2 + \frac{9}{8}y^2 - \frac{9}{4} \]

7. **Rearranging to Standard Form**:
Rearranging gives:
\[ \left( x - \frac{9}{2} \right)^2 + \frac{9}{8}y^2 = \frac{9}{4} \]
Dividing through by \( \frac{9}{4} \):
\[ \frac{\left( x - \frac{9}{2} \right)^2}{\frac{9}{4}} + \frac{\frac{9}{8}y^2}{\frac{9}{4}} = 1 \]
Simplifying gives:
\[ \frac{\left( x - \frac{9}{2} \right)^2}{\frac{9}{4}} + \frac{y^2}{2} = 1 \]

### Conclusion:
The equation we derived is in the standard form of an ellipse. Thus, the locus of the points is an ellipse.