Let the directrix of the parabola P: y^2 = 8x, cut x-axis at the point A. Let B(α, β), α > 1, be a point on P such that the slope of AB is 3/5. If BC…

For the parabola $P: y^2 = 8x$P:y2=8x, we have $4a = 8 \Rightarrow a = 2$4a=8⇒a=2.

The equation of the directrix is $x = -a \Rightarrow x = -2$x=−a⇒x=−2. Since the directrix cuts the $x$x-axis at $A$A, the coordinates of $A$A are $(-2, 0)$(−2,0).

Let the coordinates of point $B$B on the parabola be $(2t_1^2, 4t_1)$(2t12​,4t1​).

The slope of $AB$AB is given as $\dfrac{3}{5}$53​.

$$\dfrac{4t_1 - 0}{2t_1^2 - (-2)} = \dfrac{3}{5}$$2t12​−(−2)4t1​−0​=53​

$$\dfrac{2t_1}{t_1^2 + 1} = \dfrac{3}{5}$$t12​+12t1​​=53​

$$10t_1 = 3t_1^2 + 3 \Rightarrow 3t_1^2 - 10t_1 + 3 = 0$$10t1​=3t12​+3⇒3t12​−10t1​+3=0

$(3t_1 - 1)(t_1 - 3) = 0 \Rightarrow t_1 = \dfrac{1}{3}$(3t1​−1)(t1​−3)=0⇒t1​=31​ or $t_1 = 3$t1​=3.

For $t_1 = \dfrac{1}{3}$t1​=31​, $\alpha = 2\left(\dfrac{1}{3}\right)^2 = \dfrac{2}{9}$α=2(31​)2=92​, which is rejected since $\alpha > 1$α>1.

For $t_1 = 3$t1​=3, $\alpha = 2(3)^2 = 18 > 1$α=2(3)2=18>1. Thus, $B$B is $(18, 12)$(18,12).

Since $BC$BC is a focal chord, the parameter for $C$C is $t_2 = -\dfrac{1}{t_1} = -\dfrac{1}{3}$t2​=−t1​1​=−31​.

The coordinates of $C$C are $\left(2\left(-\dfrac{1}{3}\right)^2, 4\left(-\dfrac{1}{3}\right)\right) = \left(\dfrac{2}{9}, -\dfrac{4}{3}\right)$(2(−31​)2,4(−31​))=(92​,−34​).

The area of $\triangle ABC$△ABC is given by:

$$\Delta = \dfrac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$$Δ=21​∣xA​(yB​−yC​)+xB​(yC​−yA​)+xC​(yA​−yB​)∣

$$\Delta = \dfrac{1}{2} \left| -2\left(12 - \left(-\dfrac{4}{3}\right)\right) + 18\left(-\dfrac{4}{3} - 0\right) + \dfrac{2}{9}(0 - 12) \right|$$Δ=21​​−2(12−(−34​))+18(−34​−0)+92​(0−12)​

$$\Delta = \dfrac{1}{2} \left| -2\left(\dfrac{40}{3}\right) - 24 - \dfrac{24}{9} \right|$$Δ=21​​−2(340​)−24−924​​

$$\Delta = \dfrac{1}{2} \left| -\dfrac{80}{3} - 24 - \dfrac{8}{3} \right| = \dfrac{1}{2} \left| -\dfrac{88}{3} - \dfrac{72}{3} \right| = \dfrac{1}{2} \left| -\dfrac{160}{3} \right| = \dfrac{80}{3}$$Δ=21​​−380​−24−38​​=21​​−388​−372​​=21​​−3160​​=380​

Six times the area of $\triangle ABC = 6 \times \dfrac{80}{3} = 160$△ABC=6×380​=160.

Answer: $160$160