The locus of the foot of the perpendicular from the origin on each member of the family `(4a+ 3)x - (a+ 1)y -(2a+1)=0` (a) `(2x-1)^(2) +4 (y+1)^(2) = 5` (b) `(2x-1)^(2) +(y+1)^(2) = 5` (c) `(2x+1)^(2)+4(y-1)^(2) = 5` (d) `(2x-1)^(2) +4(y-1)^(2) = 5`

Options

A. `(2x-1)^(2) +4 (y+1)^(2) = 5`
B. `(2x-1)^(2) +(y+1)^(2) = 5`
C. `(2x+1)^(2)+4(y-1)^(2) = 5`
D. `(2x-1)^(2) +4(y-1)^(2) = 5`

Explanation

To find the locus of the foot of the perpendicular from the origin to the family of lines given by the equation \( (4a + 3)x - (a + 1)y - (2a + 1) = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the line
The equation of the line can be rewritten as:
\[ (4a + 3)x - (a + 1)y - (2a + 1) = 0 \]
This can be rearranged to:
\[ (4a + 3)x - (a + 1)y = (2a + 1) \]

### Step 2: Find the slope of the line
From the rearranged equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{(4a + 3)}{(a + 1)}x - \frac{(2a + 1)}{(a + 1)} \]
The slope \(m\) of the line is:
\[ m = \frac{(4a + 3)}{(a + 1)} \]

### Step 3: Find the slope of the perpendicular line
The slope of the perpendicular line from the origin will be the negative reciprocal of the slope of the line. Thus, the slope \(m_p\) of the perpendicular line is:
\[ m_p = -\frac{(a + 1)}{(4a + 3)} \]

### Step 4: Equation of the perpendicular line from the origin
The equation of the line passing through the origin with slope \(m_p\) is:
\[ y = -\frac{(a + 1)}{(4a + 3)}x \]

### Step 5: Find the intersection point
To find the foot of the perpendicular, we need to solve the equations:
1. \(y = -\frac{(a + 1)}{(4a + 3)}x\)
2. \(y = \frac{(4a + 3)}{(a + 1)}x - \frac{(2a + 1)}{(a + 1)}\)

Setting these two equations equal to each other:
\[ -\frac{(a + 1)}{(4a + 3)}x = \frac{(4a + 3)}{(a + 1)}x - \frac{(2a + 1)}{(a + 1)} \]

### Step 6: Clear the fractions
Multiply through by \((4a + 3)(a + 1)\) to eliminate the denominators:
\[ -(a + 1)^2 x = (4a + 3)^2 x - (2a + 1)(4a + 3) \]

### Step 7: Rearranging and solving for \(x\)
Rearranging gives:
\[ ((4a + 3)^2 + (a + 1)^2)x = (2a + 1)(4a + 3) \]
Thus,
\[ x = \frac{(2a + 1)(4a + 3)}{(4a + 3)^2 + (a + 1)^2} \]

### Step 8: Substitute \(x\) to find \(y\)
Substituting \(x\) back into either equation will give us \(y\). We can use the equation of the perpendicular line:
\[ y = -\frac{(a + 1)}{(4a + 3)} \cdot \frac{(2a + 1)(4a + 3)}{(4a + 3)^2 + (a + 1)^2} \]

### Step 9: Locus of the foot of the perpendicular
The coordinates \((x, y)\) will give us the locus as \(a\) varies. After simplification, we will find the equation of the locus.

### Final Result
After simplification, we find that the locus of the foot of the perpendicular from the origin is given by:
\[ (2x - 1)^2 + 4(y - 1)^2 = 5 \]

### Conclusion
Thus, the correct answer is option (d): \((2x - 1)^2 + 4(y - 1)^2 = 5\).