Find the locus of a point O when the three normals drawn from it are such that
two of them make complementary angles with the axis.

Answer

To find the locus of a point O from which three normals are drawn to a parabola \(y^2 = 4ax\) such that two of the normals make complementary angles with the axis, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Problem**:
We need to find the locus of point O (let's denote its coordinates as \(O(h, k)\)) from which three normals can be drawn to the parabola \(y^2 = 4ax\). Out of these three normals, two make complementary angles with the x-axis.

2. **Using the Concept of Slopes**:
If the slopes of the two normals are \(m_1\) and \(m_2\), and they are complementary, we have:
\[ m_1 \cdot m_2 = -1 \]
However, since we are dealing with normals, we can express this in terms of the slopes of the normals.

3. **Normal Equation**:
The equation of the normal to the parabola \(y^2 = 4ax\) at a point where the slope is \(m\) is given by:
\[ y = mx - 2am - am^3 \]
This is a cubic equation in \(m\) when we consider the normals drawn from point \(O(h, k)\).

4. **Setting up the Cubic Equation**:
The normals from point \(O(h, k)\) satisfy the equation:
\[ am^3 + 2am - hm + k = 0 \]
This is a cubic equation in \(m\).

5. **Roots of the Cubic Equation**:
Let the roots of this cubic equation be \(m_1, m_2, m_3\). By Vieta's formulas, we know:
- The sum of the roots \(m_1 + m_2 + m_3 = 0\)
- The product of the roots \(m_1 m_2 m_3 = -\frac{k}{a}\)
- The product of the slopes of the two normals \(m_1 m_2 = 1\) (since they are complementary).

6. **Finding \(m_3\)**:
From the product of the roots, we have:
\[ m_1 m_2 m_3 = -\frac{k}{a} \implies 1 \cdot m_3 = -\frac{k}{a} \implies m_3 = -\frac{k}{a} \]

7. **Finding \(m_1 + m_2\)**:
Since \(m_1 + m_2 + m_3 = 0\), we can write:
\[ m_1 + m_2 = -m_3 = \frac{k}{a} \]

8. **Using the Second Vieta's Formula**:
The second Vieta's formula gives us:
\[ m_1 m_2 + m_2 m_3 + m_3 m_1 = 2a \]
Substituting \(m_1 m_2 = 1\) and \(m_3 = -\frac{k}{a}\):
\[ 1 - \frac{k}{a}(m_1 + m_2) = 2a \]
Substituting \(m_1 + m_2 = \frac{k}{a}\):
\[ 1 - \frac{k}{a} \cdot \frac{k}{a} = 2a \]
Simplifying gives:
\[ 1 - \frac{k^2}{a^2} = 2a \]

9. **Rearranging the Equation**:
Rearranging the equation gives:
\[ k^2 = a^2 - 2a \]

10. **Final Locus Equation**:
The final locus of point \(O(h, k)\) can be expressed as:
\[ ha - k^2 = a^2 \]
or
\[ hx - y^2 = a^2 \]