Let A and B be the two points of intersection of the line $$y + 5 = 0$$ and the mirror image of the parabola $$y^{2}=4x$$ with respect to the line $$x + y + 4 = 0$$. If d denotes the distance between A and B , and a denotes the area of $$\triangle SAB$$ where $$S$$ is the focus of the parabola $$y^{2}=4x$$, then the value of (a + d) is

Reflect the line across the mirror.

Instead of reflecting the parabola, it is easier to reflect the line $$y = -5$$ across the mirror line $$L: x + y + 4 = 0$$ to see where it hits the original parabola.

The reflection of a point $$(x, y)$$ across $$x + y + c = 0$$ is $$(x', y')$$ where:

$$x' = -y - c$$ and $$y' = -x - c$$

For $$y = -5$$, any point is $$(x, -5)$$. Reflecting:

$$x' = -(-5) - 4 = 1$$

$$y' = -x - 4 \implies x = -y' - 4$$

The reflected line is $$x = 1$$.

The intersection of $$x = 1$$ and $$y^2 = 4x$$ gives $$y^2 = 4 \implies y = \pm 2$$.

Points are $$(1, 2)$$ and $$(1, -2)$$. The distance $$d$$ between these reflected points is the same as the distance between $$A$$ and $$B$$.

$$d = 2 - (-2) = 4$$.

The focus $$S$$ of $$y^2 = 4x$$ is $$(1, 0)$$.

The points $$A$$ and $$B$$ are the reflections of $$(1, 2)$$ and $$(1, -2)$$ across $$x+y+4=0$$.

• Reflection of $$(1, 2)$$: $$x = -2-4 = -6, y = -1-4 = -5 \implies A(-6, -5)$$

• Reflection of $$(1, -2)$$: $$x = 2-4 = -2, y = -1-4 = -5 \implies B(-2, -5)$$

• Area $$a$$ of $$\triangle SAB$$: $$S$$ is $$(1, 0)$$. Base $$AB = 4$$. Height = vertical distance from $$S$$ to line $$y = -5$$, which is $$0 - (-5) = 5$$.

$$a = \frac{1}{2} \times 4 \times 5 = 10$$.

Final Value: $$a + d = 10 + 4 = \mathbf{14}$$.