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

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: Understand the given information
The equation of the parabola is given as \( y^2 = 8x \). We need to find the point on this parabola where the normal line has a slope corresponding to an angle of \( 60^\circ \) with the x-axis.

**Hint:** Recall that the slope of a line inclined at an angle \( \theta \) to the x-axis is given by \( m = \tan(\theta) \).

### Step 2: Calculate the slope of the normal
The slope of the normal at the point on the parabola is given as:
\[ m_{\text{normal}} = \tan(60^\circ) = \sqrt{3} \]

**Hint:** The slope of the normal is the same as the tangent of the angle it makes with the x-axis.

### Step 3: Find the slope of the tangent
The slope of the tangent line is the negative reciprocal of the slope of the normal:
\[ m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}} = -\frac{1}{\sqrt{3}} \]

**Hint:** Remember that if the slope of the normal is \( m \), then the slope of the tangent is \( -\frac{1}{m} \).

### Step 4: Use the point-slope form of the tangent line
The equation of the tangent line to the parabola \( y^2 = 8x \) at a point \( (x_1, y_1) \) is given by:
\[ yy_1 = 4(x + x_1) \]
From this, we can express \( y \) in terms of \( x \):
\[ y = \frac{4}{y_1}x + \frac{4x_1}{y_1} \]
The slope of this line is:
\[ m = \frac{4}{y_1} \]

**Hint:** The equation of the tangent line is derived from the general form of the tangent to a parabola.

### Step 5: Set the slopes equal to find \( y_1 \)
Now, set the slope of the tangent equal to \( -\frac{1}{\sqrt{3}} \):
\[ \frac{4}{y_1} = -\frac{1}{\sqrt{3}} \]
Cross-multiplying gives:
\[ 4\sqrt{3} = -y_1 \implies y_1 = -4\sqrt{3} \]

**Hint:** Make sure to keep track of the signs when solving for \( y_1 \).

### Step 6: Substitute \( y_1 \) back into the parabola equation
Now, substitute \( y_1 \) into the parabola equation to find \( x_1 \):
\[ y_1^2 = 8x_1 \implies (-4\sqrt{3})^2 = 8x_1 \]
Calculating gives:
\[ 48 = 8x_1 \implies x_1 = \frac{48}{8} = 6 \]

**Hint:** Ensure that you square \( y_1 \) correctly and simplify the equation properly.

### Step 7: Write the coordinates of the point
Thus, the coordinates of the point on the parabola where the normal is inclined at \( 60^\circ \) to the x-axis are:
\[ (x_1, y_1) = (6, -4\sqrt{3}) \]

**Final Answer:** The coordinates are \( (6, -4\sqrt{3}) \).