From the point P(h, k) three normals are drawn to the parabola `x^(2) = 8y` and `m_(1), m_(2)` and `m_(3)` are the slopes of three normals
Find the algebraic sum of the slopes of these three normals.

Answer

To find the algebraic sum of the slopes of the three normals drawn from the point \( P(h, k) \) to the parabola \( x^2 = 8y \), we can follow these steps:

### Step 1: Understand the Parabola
The given parabola is \( x^2 = 8y \). We can rewrite it in the standard form \( y = \frac{1}{8}x^2 \).

### Step 2: Find the Slope of the Tangent
For a parabola \( y = ax^2 \), the slope of the tangent at any point \( (x_1, y_1) \) on the parabola is given by:
\[ m_t = 2ax_1 \]
In our case, \( a = \frac{1}{8} \), so the slope of the tangent becomes:
\[ m_t = 2 \cdot \frac{1}{8} \cdot x_1 = \frac{x_1}{4} \]

### Step 3: Find the Slope of the Normal
The slope of the normal \( m_n \) is the negative reciprocal of the slope of the tangent:
\[ m_n = -\frac{1}{m_t} = -\frac{4}{x_1} \]

### Step 4: Equation of the Normal
The equation of the normal line at the point \( (x_1, y_1) \) can be expressed as:
\[ y - y_1 = m_n (x - x_1) \]
Substituting \( y_1 = \frac{1}{8}x_1^2 \) and \( m_n = -\frac{4}{x_1} \):
\[ y - \frac{1}{8}x_1^2 = -\frac{4}{x_1} (x - x_1) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ y = -\frac{4}{x_1}x + \left(\frac{4}{x_1}x_1 + \frac{1}{8}x_1^2\right) \]
This simplifies to:
\[ y = -\frac{4}{x_1}x + 4 + \frac{1}{8}x_1^2 \]

### Step 6: Substitute Point \( P(h, k) \)
Since the normal passes through the point \( P(h, k) \), we substitute \( (h, k) \) into the normal equation:
\[ k = -\frac{4}{x_1}h + 4 + \frac{1}{8}x_1^2 \]
Rearranging gives:
\[ \frac{1}{8}x_1^2 - \frac{4}{x_1}h + (4 - k) = 0 \]

### Step 7: Form a Quadratic Equation
This is a quadratic equation in \( x_1 \):
\[ \frac{1}{8}x_1^2 - \frac{4h}{x_1} + (4 - k) = 0 \]
Multiplying through by \( 8x_1 \) to eliminate the fraction:
\[ x_1^3 - 32h + 8(4 - k)x_1 = 0 \]

### Step 8: Sum of the Roots
According to Vieta's formulas, for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \), the sum of the roots \( T_1 + T_2 + T_3 = -\frac{b}{a} \). Here, \( b = 0 \) (since there is no \( x^2 \) term), thus:
\[ T_1 + T_2 + T_3 = 0 \]

### Step 9: Sum of the Slopes of Normals
The slopes of the normals are given by \( m_n = -\frac{4}{T_i} \) for \( i = 1, 2, 3 \). Therefore, the sum of the slopes of the normals is:
\[ m_1 + m_2 + m_3 = -4\left(\frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}\right) \]
Using the identity for the sum of reciprocals:
\[ \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} = \frac{T_2T_3 + T_3T_1 + T_1T_2}{T_1T_2T_3} \]
From Vieta's, the product \( T_1T_2T_3 = 32h \) and the sum of products \( T_2T_3 + T_3T_1 + T_1T_2 = 4 - k \).

### Final Result
Thus, the algebraic sum of the slopes of the three normals is:
\[ m_1 + m_2 + m_3 = -4 \cdot \frac{4 - k}{32h} = \frac{k - 4}{8h} \]