From a the point P(h, k) three normals are drawn to the parabola `x^2=8y` and `m_1m_2" and "m_3` are the slopes of three normals, if the two normals from P are such that they make complementary angles with the axis then the directrix of the locus of point P (conic) is :

Options

A. `2x-3=0`
B. `2x-5=0`
C. `2y-3=0`
D. `2y-5=0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parabola and its parameters
The given parabola is \( x^2 = 8y \). We can compare this with the standard form of a parabola \( x^2 = 4ay \) to find the value of \( a \).

**Calculation:**
From \( x^2 = 4ay \), we have \( 4a = 8 \), thus \( a = 2 \).

### Step 2: Write the equation of the normal
The equation of the normal to the parabola at a point with slope \( m \) is given by:
\[ y = mx + 2a - \frac{2}{m^2} \]
Substituting \( a = 2 \):
\[ y = mx + 4 - \frac{2}{m^2} \]

### Step 3: Substitute the point \( P(h, k) \)
Since the normals are drawn from the point \( P(h, k) \), we substitute \( P(h, k) \) into the normal equation:
\[ k = mh + 4 - \frac{2}{m^2} \]
Rearranging gives:
\[ mh + \frac{2}{m^2} + k - 4 = 0 \]

### Step 4: Form a cubic equation in terms of \( m \)
Multiplying through by \( m^2 \) to eliminate the fraction:
\[ m^3 h + (k - 4)m^2 + 2 = 0 \]
This is a cubic equation in \( m \).

### Step 5: Use properties of the roots
Let \( m_1, m_2, m_3 \) be the slopes of the normals. By Vieta's formulas:
- The sum of the roots \( m_1 + m_2 + m_3 = -\frac{(k - 4)}{h} \)
- The product of the roots \( m_1 m_2 m_3 = -\frac{2}{h} \)
- The sum of the products of the roots taken two at a time \( m_1 m_2 + m_2 m_3 + m_3 m_1 = 0 \)

### Step 6: Use the condition of complementary angles
Since two normals make complementary angles, we have:
\[ m_1 m_2 = -1 \]
Let \( m_1 = t \) and \( m_2 = -\frac{1}{t} \). Then:
\[ m_3 = -\frac{2}{h} \cdot \frac{1}{m_1 m_2} = -\frac{2}{h} \cdot (-1) = \frac{2}{h} \]

### Step 7: Substitute into the sum of the slopes
Using the sum of the slopes:
\[ t - \frac{1}{t} + \frac{2}{h} = -\frac{k - 4}{h} \]
Multiplying through by \( t \) gives:
\[ t^2 - 1 + \frac{2t}{h} = -\frac{(k - 4)t}{h} \]
Rearranging leads to:
\[ ht^2 + (k - 4)t + h - 1 = 0 \]

### Step 8: Find the locus of point \( P(h, k) \)
The discriminant of this quadratic must be non-negative for real slopes:
\[ (k - 4)^2 - 4h(h - 1) \geq 0 \]
This gives the locus of \( P(h, k) \).

### Step 9: Simplify the condition
Rearranging gives:
\[ k^2 - 8k + 16 \geq 4h^2 - 4h \]
Rearranging leads to:
\[ k^2 - 8k + 16 - 4h^2 + 4h \geq 0 \]

### Step 10: Identify the directrix
The locus of point \( P \) is a parabola. The directrix can be found by comparing it with the standard form \( x^2 = 4ay \). The vertex of the parabola is at \( (0, 2) \) and the directrix is:
\[ y - 2 = -\frac{1}{2}(x - 0) \]
Thus, the directrix is:
\[ y = 3 \]

### Final Answer
The directrix of the locus of point \( P \) is:
\[ \boxed{y = 3} \]