Find the locus of a point which is such that, the three normals through it cut the axis in points whose distance from the vertex are in A.P.

Answer

To find the locus of a point such that the three normals through it cut the axis in points whose distances from the vertex are in arithmetic progression (A.P.), we can follow these steps:

### Step-by-Step Solution:

1. **Equation of the Parabola**:
The standard equation of the parabola is given by:
\[ y^2 = 4ax \]

2. **Normal to the Parabola**:
The slope of the normal at a point on the parabola can be expressed as:
\[ y = mx - 2am - am^3 \]
where \( m \) is the slope of the tangent at the point.

3. **Point on Normal**:
Let the point through which the normals pass be \( (h, k) \). The equation of the normal at this point can be rearranged to:
\[ k = mh - 2am - am^3 \]

4. **Rearranging the Normal Equation**:
We can rearrange the normal equation to:
\[ am^3 + (2a - h)m + k = 0 \]
This is a cubic equation in \( m \).

5. **Finding x-intercepts**:
To find where the normals intersect the x-axis, we set \( y = 0 \):
\[ 0 = mx - 2am - am^3 \]
Solving for \( x \):
\[ x = 2a + am^2 \]

6. **Finding x-coordinates of intercepts**:
Let the x-coordinates of the intercepts be:
\[ x_1 = 2a + am_1^2, \quad x_2 = 2a + am_2^2, \quad x_3 = 2a + am_3^2 \]

7. **Condition for A.P.**:
Since \( x_1, x_2, x_3 \) are in A.P., we have:
\[ 2x_2 = x_1 + x_3 \]

8. **Substituting x-coordinates**:
Substituting the values of \( x_1, x_2, x_3 \):
\[ 2(2a + am_2^2) = (2a + am_1^2) + (2a + am_3^2) \]
Simplifying this gives:
\[ 4a + 2am_2^2 = 4a + am_1^2 + am_3^2 \]
Cancelling \( 4a \) from both sides:
\[ 2am_2^2 = am_1^2 + am_3^2 \]

9. **Dividing by a**:
Assuming \( a \neq 0 \), we can divide by \( a \):
\[ 2m_2^2 = m_1^2 + m_3^2 \]

10. **Using properties of roots**:
From the cubic equation \( am^3 + (2a - h)m + k = 0 \), we know:
- Sum of roots \( m_1 + m_2 + m_3 = 0 \)
- Sum of products of roots taken two at a time \( m_1m_2 + m_2m_3 + m_3m_1 = \frac{2a - h}{a} \)
- Product of roots \( m_1m_2m_3 = -\frac{k}{a} \)

11. **Substituting into the A.P. condition**:
We can express \( m_1^2 + m_3^2 \) in terms of \( m_2 \):
\[ m_1^2 + m_3^2 = (m_1 + m_3)^2 - 2m_1m_3 \]
Since \( m_1 + m_3 = -m_2 \):
\[ m_1^2 + m_3^2 = m_2^2 - 2m_1m_3 \]

12. **Final Equation**:
Substituting back into the A.P. condition gives us:
\[ 2m_2^2 = m_2^2 - 2m_1m_3 \]
Rearranging leads to:
\[ m_2^2 + 2m_1m_3 = 0 \]

13. **Finding the locus**:
Using the relationships derived, we can express the locus in terms of \( h \) and \( k \):
\[ 27k^2 = 2a - h \]
Rearranging gives us the final locus:
\[ 27ay^2 = 2a - x \]

### Final Answer:
The locus of the point is given by:
\[ 27ay^2 = 2a - x \]