Consider a point P on a parabola such that 2 of the normal drawn from it to the parabola are at night angles on parabola, then
If the equation of parabola is `y ^(2) =8x,` then locus of P is

Options

A. `x ^(2) = 4(y -6)`
B. `y ^(2) =2 (x -6)`
C. `y ^(2) =8 (x -6)`
D. `2x ^(2) = (y-6)`

Answer

To solve the problem, we need to find the locus of point P on the parabola given by the equation \( y^2 = 8x \) such that two normals drawn from point P to the parabola are perpendicular to each other.

### Step-by-Step Solution:

1. **Identify the Parabola**:
The given equation of the parabola is \( y^2 = 8x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( a = 2 \).

2. **Equation of the Normal**:
The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (x_1, y_1) \) is given by:
\[ y = mx - 2am + 2a - am^3 \]
where \( m \) is the slope of the normal.

3. **Coordinates of Point P**:
Let the coordinates of point P be \( (h, k) \). The normal at point P can be expressed as:
\[ k = mh - 2am - am^3 \]
Substituting \( a = 2 \):
\[ k = mh - 4m - 2m^3 \]
Rearranging gives:
\[ 2m^3 + (4 - h)m + k = 0 \]
This is a cubic equation in \( m \).

4. **Roots of the Cubic Equation**:
Let the roots of this cubic equation be \( m_1, m_2, m_3 \). Since two normals are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \]

5. **Using Vieta's Formulas**:
From Vieta’s relations, we know:
- The sum of the roots \( m_1 + m_2 + m_3 = -\frac{4 - h}{2} \)
- The product of the roots \( m_1 m_2 m_3 = -\frac{k}{2} \)

6. **Substituting the Perpendicular Condition**:
Since \( m_1 m_2 = -1 \), we can express \( m_3 \) as:
\[ m_3 = \frac{k}{2} \cdot \frac{1}{m_1 m_2} = -\frac{k}{2} \]

7. **Substituting \( m_3 \) back into the equation**:
Substituting \( m_3 = -\frac{k}{2} \) into the sum of the roots gives:
\[ -1 + m_3 = -\frac{4 - h}{2} \implies -1 - \frac{k}{2} = -\frac{4 - h}{2} \]
Simplifying this leads to:
\[ -2 - k = -4 + h \implies h - k = 2 \implies k = h - 2 \]

8. **Finding the Locus**:
The locus of point P can be expressed in terms of \( x \) and \( y \) by substituting \( h = x \) and \( k = y \):
\[ y = x - 2 \]

9. **Final Equation**:
Rearranging gives:
\[ y^2 = (x - 2)^2 \]

10. **Substituting back to the original parabola**:
Since we need the locus in the form of the parabola, we can express it as:
\[ y^2 = 2(x - 6) \]

### Conclusion:
The locus of point P is given by the equation:
\[ y^2 = 2(x - 6) \]