The point on the parabola `y^(2) = 8x` at which the normal is inclined at `60^(@)` to the x-axis has the co-ordinates as (a) `(6,-4sqrt(3))` (b) `(6,4sqrt(3))` (c) `(-6,-4sqrt(3))` (d) `(-6,4sqrt(3))`

Options

A. `(6,-4sqrt(3))`
B. `(6,4sqrt(3))`
C. `(-6,-4sqrt(3))`
D. `(-6,4sqrt(3))`

Explanation

To find the point on the parabola \( y^2 = 8x \) where the normal is inclined at \( 60^\circ \) to the x-axis, we can follow these steps:

### Step 1: Determine the slope of the normal
The slope of a line inclined at an angle \( \theta \) to the x-axis is given by \( \tan(\theta) \). For \( \theta = 60^\circ \):
\[ M_1 = \tan(60^\circ) = \sqrt{3} \]
Thus, the slope of the normal \( M_1 = \sqrt{3} \).

### Step 2: Find the slope of the tangent
Since the normal and tangent are perpendicular, the slope of the tangent \( M_2 \) can be found using the relationship:
\[ M_1 \cdot M_2 = -1 \]
Therefore:
\[ M_2 = -\frac{1}{M_1} = -\frac{1}{\sqrt{3}} \]

### Step 3: Use the equation of the parabola
The equation of the parabola is given as \( y^2 = 8x \). We can rewrite this in the standard form \( y^2 = 4ax \) where \( 4a = 8 \) implies \( a = 2 \).

### Step 4: Write the equation of the tangent
The equation of the tangent to the parabola at the point \( (x_1, y_1) \) is given by:
\[ yy_1 = 4(x + x_1) \]
Substituting \( a = 2 \):
\[ yy_1 = 4x + 4x_1 \]

### Step 5: Determine the slope of the tangent
The slope of the tangent line can also be expressed as:
\[ \text{slope} = \frac{4}{y_1} \]
Setting this equal to the slope we found earlier:
\[ \frac{4}{y_1} = -\frac{1}{\sqrt{3}} \]

### Step 6: Solve for \( y_1 \)
Cross-multiplying gives:
\[ 4\sqrt{3} = -y_1 \quad \Rightarrow \quad y_1 = -4\sqrt{3} \]

### Step 7: Substitute \( y_1 \) back into the parabola
Now we substitute \( y_1 \) back into the parabola equation to find \( x_1 \):
\[ y_1^2 = 8x_1 \quad \Rightarrow \quad (-4\sqrt{3})^2 = 8x_1 \]
Calculating \( y_1^2 \):
\[ 16 \cdot 3 = 48 = 8x_1 \quad \Rightarrow \quad x_1 = \frac{48}{8} = 6 \]

### Step 8: Final coordinates
Thus, the coordinates of the point on the parabola are:
\[ (x_1, y_1) = (6, -4\sqrt{3}) \]

### Conclusion
The correct answer is:
**(a) \( (6, -4\sqrt{3}) \)**

---