Three normals to the parabola `y^2 = x` are drawn through a point (c,0), then

Options

A. `c=1//4`
B. `c=1//2`
C. `c gt 1//2`
D. none of these

Answer

To solve the problem, we need to find the condition on \( c \) such that three normals to the parabola \( y^2 = x \) can be drawn through the point \( (c, 0) \).

### Step-by-Step Solution:

1. **Identify the Parabola**: The given parabola is \( y^2 = x \). This can be compared to the standard form \( y^2 = 4ax \), where \( a = 1/4 \).

2. **Equation of the Normal**: The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ y = mx - 2am - am^3 \]
Here, \( m \) is the slope of the normal.

3. **Using the Point (c, 0)**: Since the normal passes through the point \( (c, 0) \), we substitute \( y = 0 \) and \( x = c \) into the normal equation:
\[ 0 = mc - 2am - am^3 \]
Rearranging gives:
\[ mc - 2am - am^3 = 0 \]
This can be factored as:
\[ m(c - 2a - am^2) = 0 \]
Since \( m \neq 0 \), we have:
\[ c - 2a - am^2 = 0 \]
Thus,
\[ c = 2a + am^2 \]

4. **Substituting for \( a \)**: We know \( a = \frac{1}{4} \), so:
\[ c = 2 \cdot \frac{1}{4} + \frac{1}{4}m^2 = \frac{1}{2} + \frac{1}{4}m^2 \]

5. **Finding the Condition on \( c \)**: Rearranging gives:
\[ m^2 = 4c - 2 \]
Since \( m^2 \geq 0 \), we have:
\[ 4c - 2 \geq 0 \]
This simplifies to:
\[ 4c \geq 2 \quad \Rightarrow \quad c \geq \frac{1}{2} \]

6. **Conclusion**: The condition for \( c \) such that three normals can be drawn to the parabola from the point \( (c, 0) \) is:
\[ c > \frac{1}{2} \]