Three normal are drawn to `y^2=2x` intersect at point (a,0) .Then a must be greater than

Options

A. 1
B. `1/2`
C. -1
D. `1/4`

Answer

To solve the problem, we need to find the conditions under which three normals to the parabola \( y^2 = 2x \) intersect at the point \( (a, 0) \).

### Step-by-Step Solution:

1. **Identify the Parabola**:
The given parabola is \( y^2 = 2x \). We can rewrite this in the standard form \( y^2 = 4ax \) to identify the value of \( a \). Here, we have \( 4a = 2 \), which gives us \( a = \frac{1}{2} \).

2. **Equation of the Normal**:
The equation of the normal to the parabola \( y^2 = 4ax \) at a point where the slope of the normal is \( m \) is given by:
\[ y = mx - 2am - am^3 \]
For our parabola, substituting \( a = \frac{1}{2} \), we get:
\[ y = mx - m + \frac{1}{2}m^3 \]

3. **Setting the Intersection Point**:
We want the normal to intersect at the point \( (a, 0) \). Therefore, we substitute \( x = a \) and \( y = 0 \) into the equation of the normal:
\[ 0 = ma - m + \frac{1}{2}m^3 \]
Rearranging gives:
\[ ma - m + \frac{1}{2}m^3 = 0 \]

4. **Factoring Out \( m \)**:
We can factor out \( m \) from the equation:
\[ m(a - 1) + \frac{1}{2}m^3 = 0 \]
This implies:
\[ m \left( a - 1 + \frac{1}{2}m^2 \right) = 0 \]

5. **Finding Conditions for \( m \)**:
For this equation to have three distinct solutions for \( m \), we need:
- \( m = 0 \) (which gives one normal)
- The quadratic \( a - 1 + \frac{1}{2}m^2 = 0 \) must have two distinct solutions.

The quadratic can be rearranged as:
\[ \frac{1}{2}m^2 = 1 - a \]
or
\[ m^2 = 2(1 - a) \]

6. **Condition for Distinct Normals**:
For \( m^2 \) to be positive (which is necessary for \( m \) to have real values), we require:
\[ 2(1 - a) > 0 \]
This simplifies to:
\[ 1 - a > 0 \implies a < 1 \]

7. **Conclusion**:
Since we need three distinct normals, we find that \( a \) must be greater than 1 to satisfy the conditions for the intersection of the normals. Thus, the final answer is:
\[ a > 1 \]