Three normals are drawn from the point (a,0) to the parabola `y^2=x`. One normal is the X-axis . If other two normals are perpendicular to each other , then the value of 4a is

Answer

To solve the problem, we need to find the value of \(4a\) given that three normals are drawn from the point \((a, 0)\) to the parabola \(y^2 = x\). One of the normals is the x-axis, and the other two normals are perpendicular to each other.

### Step-by-step Solution:

1. **Understanding the Parabola**: The equation of the parabola is given by \(y^2 = x\). This can be rewritten in the standard form \(y^2 = 4Ax\) where \(4A = 1\), thus \(A = \frac{1}{4}\).

2. **Equation of the Normal**: The equation of the normal to the parabola \(y^2 = 4Ax\) at a point where the slope is \(m\) is given by:
\[ y = mx - 2A - \frac{A m^3}{4} \]
Substituting \(A = \frac{1}{4}\), we get:
\[ y = mx - \frac{1}{2} - \frac{m^3}{16} \]

3. **Point of Intersection**: Since the normal passes through the point \((a, 0)\), we substitute \(y = 0\) and \(x = a\) into the normal equation:
\[ 0 = ma - \frac{1}{2} - \frac{m^3}{16} \]
Rearranging gives:
\[ ma = \frac{1}{2} + \frac{m^3}{16} \]

4. **Rearranging the Equation**: We can express this as:
\[ ma - \frac{m^3}{16} = \frac{1}{2} \]
Factoring out \(m\):
\[ m\left(a - \frac{m^2}{16}\right) = \frac{1}{2} \]

5. **Finding the Slopes**: We can find the slopes of the normals. Let \(m_1 = 0\) (for the x-axis) and the other two slopes be \(m_2 = 2\sqrt{a - \frac{1}{2}}\) and \(m_3 = -2\sqrt{a - \frac{1}{2}}\).

6. **Condition for Perpendicularity**: The normals are perpendicular if:
\[ m_2 \cdot m_3 = -1 \]
Substituting the values gives:
\[ (2\sqrt{a - \frac{1}{2}})(-2\sqrt{a - \frac{1}{2}}) = -1 \]
Simplifying:
\[ -4(a - \frac{1}{2}) = -1 \]
Thus:
\[ 4(a - \frac{1}{2}) = 1 \]

7. **Solving for \(a\)**: Rearranging gives:
\[ 4a - 2 = 1 \implies 4a = 3 \]

8. **Final Answer**: Therefore, the value of \(4a\) is:
\[ \boxed{3} \]