Let the directrix of the parabola P : y^2 = 8 | vedclass

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(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$, we get $a = 2$.The directrix is $x = -a$, so $x = -2$. The point $A$ where

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$160$

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(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$, we get $a = 2$.The directrix is $x = -a$, so $x = -2$. The point $A$ where the directrix cuts the $x$-axis is $A(-2, 0)$.Point $B(\alpha, \beta)$ lies on $y^2 = 8x$, so $\beta^2 = 8\alpha$.The slope of $AB$ is given by $\frac{\beta - 0}{\alpha - (-2)} = \frac{\beta}{\alpha + 2} = \frac{3}{5}$.Thus, $5\beta = 3\alpha + 6$. Substituting $\alpha = \frac{\beta^2}{8}$, we get $5\beta = 3(\frac{\beta^2}{8}) + 6$, which simplifies to $3\beta^2 - 40\beta + 48 = 0$.Factoring gives $(3\beta - 4)(\beta - 12) = 0$. Since $\alpha > 1$, $\beta^2 = 8\alpha > 8$, so $\beta > 2.82$. Thus, $\beta = 12$ and $\alpha = 18$. So $B = (18, 12)$.The focal chord $BC$ passes through the focus $S(2, 0)$. The line $BC$ passes through $(18, 12)$ and $(2, 0)$.The slope of $BC$ is $m = \frac{12 - 0}{18 - 2} = \frac{12}{16} = \frac{3}{4}$.The equation of $BC$ is $y - 0 = \frac{3}{4}(x - 2) \Rightarrow y = \frac{3}{4}x - \frac{3}{2}$.Substituting into $y^2 = 8x$: $(\frac{3}{4}x - \frac{3}{2})^2 = 8x \Rightarrow \frac{9}{16}(x-2)^2 = 8x \Rightarrow 9(x^2 - 4x + 4) = 128x \Rightarrow 9x^2 - 164x + 36 = 0$.Since $x_B = 18$ is a root, $x_C = \frac{36}{9 	imes 18} = \frac{2}{9}$.Then $y_C = \frac{3}{4}(\frac{2}{9}) - \frac{3}{2} = \frac{1}{6} - \frac{9}{6} = -\frac{8}{6} = -\frac{4}{3}$.Area of $	riangle ABC = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = \frac{1}{2} |-2(12 - (-4/3)) + 18(-4/3 - 0) + 2/9(0 - 12)| = \frac{1}{2} |-2(40/3) - 24 - 8/3| = \frac{1}{2} |-80/3 - 72/3 - 8/3| = \frac{1}{2} |-160/3| = 80/3$.Six times the area $= 6 	imes (80/3) = 160$.

Let the directrix of the parabola $P : y^2 = 8x$, cut $x$-axis at the point $A$. Let $B(\alpha, \beta)$, $\alpha > 1$, be a point on $P$ such that the slope of $AB$ is $3/5$. If $BC$ is a focal chord of $P$, then six times the area of $\triangle ABC$ is :

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