A focal chord for parabola `y^(2)=8(x+2)` is inclined at an angle of `60^(@)` with positive x-axis and intersects the parabola at P and Q. Let perpendicular bisector of the chord PQ intersects the x-axis at R, then the distance of R from focus is :

Options

A. `(8)/(3)`
B. `(16sqrt(3))/(3)`
C. `(16)/(3)`
D. `8sqrt(3)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the Parabola and its Properties
The given parabola is \( y^2 = 8(x + 2) \).

**Hint:** The standard form of a parabola is \( y^2 = 4ax \). Here, \( 4a = 8 \) implies \( a = 2 \). The vertex of the parabola is at \( (-2, 0) \) and the focus is at \( (0, 0) \).

### Step 2: Determine the Slope of the Focal Chord
The focal chord is inclined at an angle of \( 60^\circ \) with the positive x-axis. The slope \( m \) of the line can be calculated as:
\[ m = \tan(60^\circ) = \sqrt{3} \]

**Hint:** Remember that the slope of a line is given by the tangent of the angle it makes with the positive x-axis.

### Step 3: Write the Equation of the Focal Chord
Using the slope, the equation of the line passing through the focus \( (0, 0) \) is:
\[ y = \sqrt{3}x \]

**Hint:** The equation of a line through the origin with slope \( m \) is \( y = mx \).

### Step 4: Find the Points of Intersection with the Parabola
Substituting \( y = \sqrt{3}x \) into the parabola's equation:
\[ (\sqrt{3}x)^2 = 8(x + 2) \]
This simplifies to:
\[ 3x^2 = 8x + 16 \]
Rearranging gives:
\[ 3x^2 - 8x - 16 = 0 \]

**Hint:** This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \).

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -8 \), \( c = -16 \).
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ x = \frac{8 \pm \sqrt{64 + 192}}{6} = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6} \]
This gives us:
\[ x_1 = 4, \quad x_2 = -\frac{4}{3} \]

**Hint:** Ensure to simplify the square root and perform the arithmetic correctly.

### Step 6: Find Corresponding y-coordinates
Using \( y = \sqrt{3}x \):
- For \( x_1 = 4 \):
\[ y_1 = \sqrt{3} \cdot 4 = 4\sqrt{3} \]
- For \( x_2 = -\frac{4}{3} \):
\[ y_2 = \sqrt{3} \cdot \left(-\frac{4}{3}\right) = -\frac{4\sqrt{3}}{3} \]

**Hint:** Substitute back into the line equation to find the y-coordinates.

### Step 7: Find the Midpoint of PQ
The points \( P(4, 4\sqrt{3}) \) and \( Q\left(-\frac{4}{3}, -\frac{4\sqrt{3}}{3}\right) \) give the midpoint \( M \):
\[ M_x = \frac{4 - \frac{4}{3}}{2} = \frac{\frac{12}{3} - \frac{4}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{4}{3} \]
\[ M_y = \frac{4\sqrt{3} - \left(-\frac{4\sqrt{3}}{3}\right)}{2} = \frac{4\sqrt{3} + \frac{4\sqrt{3}}{3}}{2} = \frac{\frac{12\sqrt{3}}{3} + \frac{4\sqrt{3}}{3}}{2} = \frac{\frac{16\sqrt{3}}{3}}{2} = \frac{8\sqrt{3}}{3} \]

**Hint:** The midpoint formula is \( M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).

### Step 8: Find the Perpendicular Bisector
The slope of the line segment \( PQ \) is:
\[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{4\sqrt{3}}{3} - 4\sqrt{3}}{-\frac{4}{3} - 4} = \frac{-\frac{16\sqrt{3}}{3}}{-\frac{16}{3}} = \sqrt{3} \]
The slope of the perpendicular bisector is:
\[ m_{perpendicular} = -\frac{1}{\sqrt{3}} \]
Using the midpoint \( M\left(\frac{4}{3}, \frac{8\sqrt{3}}{3}\right) \), the equation is:
\[ y - \frac{8\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}\left(x - \frac{4}{3}\right) \]

**Hint:** The slope of the perpendicular line is the negative reciprocal of the original slope.

### Step 9: Find Intersection with the X-axis
Set \( y = 0 \) in the equation of the perpendicular bisector and solve for \( x \):
\[ 0 - \frac{8\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}\left(x - \frac{4}{3}\right) \]
Solving gives:
\[ x = \frac{16}{3} \]

**Hint:** Substitute \( y = 0 \) to find where the line intersects the x-axis.

### Step 10: Calculate the Distance from Focus
The focus is at \( (0, 0) \) and the point \( R\left(\frac{16}{3}, 0\right) \):
\[ \text{Distance} = \left| \frac{16}{3} - 0 \right| = \frac{16}{3} \]

**Hint:** The distance between two points on the x-axis is simply the absolute difference of their x-coordinates.

### Final Answer
The distance of \( R \) from the focus is \( \frac{16}{3} \).