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 _____.
Step-by-Step Solution
Key Concept: Find the mirror image of a general point $P(t^2,2t)$ on $y^2=4x$ across $x+y+4=0$; eliminate $t$ to get the image parabola; then intersect with $y=-5$ to find $A$ and $B$.
Reflecting $P(t^2,2t)$ in $x+y+4=0$: the signed distance is $-(t^2+2t+4)$, giving $x'=t^2-2(t+1)^2+...$; the result is $x'=-2t-4$, $y'=-t^2-4$. Eliminating $t$: $t=\tfrac{-x'-4}{2}$, so $y'+4=-(x'+4)^2/4$, i.e., $(x+4)^2=-4(y+4)$ — a downward parabola with focus $(-4,-5)$. Intersecting with $y=-5$: $(x+4)^2=4$, giving $x=-2$ or $x=-6$. So $A=(-2,-5)$, $B=(-6,-5)$, $d=4$. Focus of $y^2=4x$ is $S=(1,0)$. Area $a=\tfrac{1}{2}\cdot d\cdot|y_S-y_{AB}|=\tfrac{1}{2}\cdot4\cdot|0-(-5)|=10$. Hence $a+d=14$.
Correct Answer: 14