Find the locus of a point , which moves such that its distance from the point (0,-1) is twice its distance from the line 3x+4y+1=0.

Answer

To find the locus of a point that moves such that its distance from the point (0, -1) is twice its distance from the line \(3x + 4y + 1 = 0\), we can follow these steps:

### Step 1: Define the point
Let the moving point be \(P(p, q)\).

### Step 2: Calculate the distance from the point (0, -1)
The distance \(d_1\) from the point \(P(p, q)\) to the point \((0, -1)\) is given by the distance formula:
\[ d_1 = \sqrt{(p - 0)^2 + (q + 1)^2} = \sqrt{p^2 + (q + 1)^2} \]

### Step 3: Calculate the distance from the line
The distance \(d_2\) from the point \(P(p, q)\) to the line \(3x + 4y + 1 = 0\) can be calculated using the formula for the distance from a point to a line:
\[ d_2 = \frac{|3p + 4q + 1|}{\sqrt{3^2 + 4^2}} = \frac{|3p + 4q + 1|}{5} \]

### Step 4: Set up the relationship between distances
According to the problem, the distance from the point to (0, -1) is twice the distance from the point to the line:
\[ \sqrt{p^2 + (q + 1)^2} = 2 \cdot \frac{|3p + 4q + 1|}{5} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ p^2 + (q + 1)^2 = \left(2 \cdot \frac{3p + 4q + 1}{5}\right)^2 \]
\[ p^2 + (q + 1)^2 = \frac{4(3p + 4q + 1)^2}{25} \]

### Step 6: Expand both sides
Expanding the left side:
\[ p^2 + (q^2 + 2q + 1) = p^2 + q^2 + 2q + 1 \]

Expanding the right side:
\[ \frac{4(9p^2 + 24pq + 16q^2 + 6p + 8q + 1)}{25} \]
\[ = \frac{36p^2 + 96pq + 64q^2 + 24p + 32q + 4}{25} \]

### Step 7: Multiply through by 25 to eliminate the fraction
\[ 25(p^2 + q^2 + 2q + 1) = 36p^2 + 96pq + 64q^2 + 24p + 32q + 4 \]
\[ 25p^2 + 25q^2 + 50q + 25 = 36p^2 + 96pq + 64q^2 + 24p + 32q + 4 \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ 0 = 36p^2 - 25p^2 + 96pq + 64q^2 - 25q^2 + 24p + 32q - 50q + 4 - 25 \]
\[ 0 = 11p^2 + 32pq + 39q^2 + 24p - 18q - 21 \]

### Step 9: Replace \(p\) and \(q\) with \(x\) and \(y\)
Finally, we replace \(p\) and \(q\) with \(x\) and \(y\):
\[ 11x^2 + 32xy + 39y^2 + 24x - 18y - 21 = 0 \]

### Final Answer
The locus of the point is given by the equation:
\[ 11x^2 + 32xy + 39y^2 + 24x - 18y - 21 = 0 \]