Six Times Area of Triangle Formed by Directrix Intersection and Focal Chord

To find six times the area of $\Delta ABC$, we proceed step-by-step:

Step 1: Identify the coordinates of point $A$

The given parabola is $P : y^2 = 8x$.
Comparing this with the standard equation of a parabola $y^2 = 4ax$, we get $4a = 8 \implies a = 2$.

The directrix of the parabola is given by the equation: $x = -a \implies x = -2$

Since point $A$ is the intersection of the directrix and the $x$-axis ($y = 0$), the coordinates of $A$ are: $A = (-2, 0)$

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Step 2: Determine the coordinates of point $B$

Let $B = (\alpha, \beta)$ be a point on the parabola $P$. Thus, it satisfies: $\beta^2 = 8\alpha$

The slope of the line segment $AB$ is given as $\frac{3}{5}$: $\text{Slope}(AB) = \frac{\beta - 0}{\alpha - (-2)} = \frac{3}{5} \implies 5\beta = 3(\alpha + 2)$

Substitute $\alpha = \frac{\beta^2}{8}$ into the equation: $3\left(\frac{\beta^2}{8}\right) - 5\beta + 6 = 0$ $3\beta^2 - 40\beta + 48 = 0$

Factoring the quadratic equation: $(3\beta - 4)(\beta - 12) = 0$

This gives two possible values for $\beta$:

1. If $\beta = \frac{4}{3}$, then $\alpha = \frac{(4/3)^2}{8} = \frac{2}{9}$. However, the problem states that $\alpha > 1$, so this solution is rejected.
2. If $\beta = 12$, then $\alpha = \frac{12^2}{8} = 18$, which satisfies $\alpha > 1$.

Thus, the coordinates of point $B$ are: $B = (18, 12)$

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Step 3: Determine the coordinates of point $C$

The parametric coordinates for any point on the parabola $y^2 = 4ax$ with $a = 2$ are given by $(at^2, 2at) = (2t^2, 4t)$.

For point $B(18, 12)$: $4t_1 = 12 \implies t_1 = 3$

Since $BC$ is a focal chord, the parameter $t_2$ of point $C$ satisfies the standard relation $t_1 t_2 = -1$: $3 t_2 = -1 \implies t_2 = -\frac{1}{3}$

Substituting $t_2 = -\frac{1}{3}$ into the parametric form gives the coordinates of $C$: $x_C = 2t_2^2 = 2\left(-\frac{1}{3}\right)^2 = \frac{2}{9}$ $y_C = 4t_2 = 4\left(-\frac{1}{3}\right) = -\frac{4}{3}$

Thus, the coordinates of point $C$ are: $C = \left(\frac{2}{9}, -\frac{4}{3}\right)$

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Step 4: Calculate the Area of $\Delta ABC$

Using the shoelace formula for the vertices $A(-2, 0)$, $B(18, 12)$, and $C\left(\frac{2}{9}, -\frac{4}{3}\right)$:

$\text{Area}(\Delta ABC) = \frac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$

Substitute the values: $\text{Area}(\Delta ABC) = \frac{1}{2} \left| -2\left(12 - \left(-\frac{4}{3}\right)\right) + 18\left(-\frac{4}{3} - 0\right) + \frac{2}{9}(0 - 12) \right|$ $\text{Area}(\Delta ABC) = \frac{1}{2} \left| -2\left(\frac{40}{3}\right) - 24 - \frac{8}{3} \right|$ $\text{Area}(\Delta ABC) = \frac{1}{2} \left| -\frac{80}{3} - \frac{72}{3} - \frac{8}{3} \right|$ $\text{Area}(\Delta ABC) = \frac{1}{2} \left| -\frac{160}{3} \right| = \frac{80}{3}$

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Step 5: Calculate Six Times the Area

$6 \times \text{Area}(\Delta ABC) = 6 \times \frac{80}{3} = 160$

Hence, the correct option is B.