Consider a point P on a parabola such that 2 of the normal drawn from it to the parabola are at right angles on parabola, then
If `P -= (x _(1), y _(1)),` the slope of third normal is, if If the equation of parabola is `y^(2)= 8x`

Options

A. `(y _(1))/(8)`
B. `(y _(1))/(2)`
C. `-(y _(1))/(8)`
D. `- (y _(1))/(2)`

Answer

To find the slope of the third normal from a point \( P(x_1, y_1) \) on the parabola given by the equation \( y^2 = 8x \), we can follow these steps:

### Step 1: Understand the equation of the parabola
The given parabola is \( y^2 = 8x \). This can be rewritten in the standard form of a parabola, which is \( y^2 = 4px \), where \( p = 2 \).

### Step 2: Find the equation of the normal
The slope of the tangent line at a point \( (x_0, y_0) \) on the parabola can be derived from the derivative. For the parabola \( y^2 = 8x \):
- Differentiate implicitly: \( 2y \frac{dy}{dx} = 8 \) → \( \frac{dy}{dx} = \frac{4}{y} \).
- At point \( P(x_1, y_1) \), the slope of the tangent (let's denote it as \( m_t \)) is \( \frac{4}{y_1} \).

The slope of the normal line (denote it as \( m_n \)) is the negative reciprocal of the slope of the tangent:
\[ m_n = -\frac{y_1}{4} \]

### Step 3: Write the equation of the normal line
Using the point-slope form of the line, the equation of the normal line at point \( P(x_1, y_1) \) is:
\[ y - y_1 = m_n(x - x_1) \]
Substituting \( m_n \):
\[ y - y_1 = -\frac{y_1}{4}(x - x_1) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ y = -\frac{y_1}{4}x + \left(\frac{y_1}{4}x_1 + y_1\right) \]

### Step 5: Finding the cubic equation for slopes
From the normal line equation, we can express \( y \) in terms of \( x \) and \( m \):
\[ y = mx - 4m - 2m^3 \]
Substituting \( (x_1, y_1) \) into this equation gives:
\[ y_1 = mx_1 - 4m - 2m^3 \]
Rearranging leads to:
\[ 2m^3 + (4 - x_1)m + y_1 = 0 \]

### Step 6: Roots of the cubic equation
Let the roots of the cubic equation be \( m_1, m_2, m_3 \). We know that for two normals to be perpendicular, their slopes must satisfy:
\[ m_1 \cdot m_2 = -1 \]

### Step 7: Using Vieta's formulas
From Vieta's formulas, we know:
\[ m_1 + m_2 + m_3 = -\frac{4 - x_1}{2} \]
\[ m_1 m_2 + m_2 m_3 + m_3 m_1 = 0 \]
\[ m_1 m_2 m_3 = -\frac{y_1}{2} \]

### Step 8: Finding the slope of the third normal
Since \( m_1 m_2 = -1 \):
\[ m_3 = -\frac{y_1}{2} \]

### Final Result
Thus, the slope of the third normal is:
\[ m_3 = \frac{y_1}{2} \]