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 \( P(h, k) \) 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: Distance from the Point
First, we calculate the distance from the point \( (0, -1) \) to the point \( P(h, k) \):
\[ d_1 = \sqrt{(h - 0)^2 + (k + 1)^2} = \sqrt{h^2 + (k + 1)^2} \]

### Step 2: Distance from the Line
Next, we calculate the distance from the point \( P(h, k) \) to the line \( 3x + 4y + 1 = 0 \). The formula for the distance from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d_2 = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, \( A = 3 \), \( B = 4 \), and \( C = 1 \):
\[ d_2 = \frac{|3h + 4k + 1|}{\sqrt{3^2 + 4^2}} = \frac{|3h + 4k + 1|}{5} \]

### Step 3: Set Up the Equation
According to the problem, the distance from the point \( (0, -1) \) is twice the distance from the line:
\[ \sqrt{h^2 + (k + 1)^2} = 2 \cdot \frac{|3h + 4k + 1|}{5} \]

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

### Step 5: Expand Both Sides
Now we expand both sides:
1. Left side:
\[ h^2 + (k^2 + 2k + 1) = h^2 + k^2 + 2k + 1 \]
2. Right side:
\[ \frac{4(9h^2 + 24hk + 16k^2 + 6h + 8k + 1)}{25} = \frac{36h^2 + 96hk + 64k^2 + 24h + 32k + 4}{25} \]

### Step 6: Clear the Denominator
Multiply through by 25 to eliminate the fraction:
\[ 25(h^2 + k^2 + 2k + 1) = 4(9h^2 + 24hk + 16k^2 + 6h + 8k + 1) \]
This simplifies to:
\[ 25h^2 + 25k^2 + 50k + 25 = 36h^2 + 96hk + 64k^2 + 24h + 32k + 4 \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ 0 = 36h^2 - 25h^2 + 64k^2 - 25k^2 + 96hk + 24h - 50k + 4 - 25 \]
\[ 0 = 11h^2 + 39k^2 + 96hk + 24h - 18k - 21 \]

### Step 8: Factor Out Common Terms
We can factor out a common term of 3:
\[ 0 = 3(11h^2 + 13k^2 + 32hk + 8h - 6k - 7) \]

### Step 9: Replace Variables
Finally, replace \( h \) and \( k \) with \( x \) and \( y \):
\[ 11x^2 + 13y^2 + 32xy + 8x - 6y - 7 = 0 \]

### Final Answer
Thus, the required locus is:
\[ 11x^2 + 13y^2 + 32xy + 8x - 6y - 7 = 0 \]