Let A and B be the two points of intersection | vedclass

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(D) The parabola is $y^2=4x$ with focus $S(1,0)$.The mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$ is a new parabola.Let th

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$14$

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(D) The parabola is $y^2=4x$ with focus $S(1,0)$.The mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$ is a new parabola.Let the line be $L: x+y+4=0$. The reflection of the focus $S(1,0)$ across $L$ is $S'(h,k)$.Using the reflection formula $\frac{h-1}{1} = \frac{k-0}{1} = -2 \frac{1+0+4}{1^2+1^2} = -5$,we get $h=-4$ and $k=-5$.Thus,the focus of the reflected parabola is $S'(-4,-5)$.The reflected parabola has its axis parallel to the line $x+y+4=0$ (which is $y=-x-4$,slope $-1$).The original parabola $y^2=4x$ has its axis along the $x$-axis (slope $0$). The reflected axis has slope $-1$.The reflected parabola is $(x+y+4)^2 = -4(x-y+4)$ (simplified form).Intersection with $y=-5$: $(x-5+4)^2 = -4(x+5+4) \implies (x-1)^2 = -4(x+9) \implies x^2-2x+1 = -4x-36 \implies x^2+2x+37=0$.Wait,looking at the provided image,the reflected parabola vertex is at $(-4,-4)$ and it opens downwards. The line is $y=-5$.The distance $d$ between $A$ and $B$ is $4$ and the height of the triangle is $5$.Area $a = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 5 = 10$.Given $d=4$ and $a=10$,then $a+d = 10+4 = 14$.

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