Let the curve C be the mirror image of the pa | vedclass

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(A) The mirror image of the line $y = -5$ with respect to the line $x + y + 4 = 0$ is found by reflecting the line.Let a point on $y = -5$ be $(h, -5)

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$4$

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(A) The mirror image of the line $y = -5$ with respect to the line $x + y + 4 = 0$ is found by reflecting the line.Let a point on $y = -5$ be $(h, -5)$. The reflection $(x', y')$ of $(h, -5)$ across $x + y + 4 = 0$ satisfies $\frac{x' - h}{1} = \frac{y' + 5}{1} = -2 \frac{h - 5 + 4}{1^2 + 1^2} = -(h - 1)$.Thus,$x' = h - (h - 1) = 1$ and $y' = -5 - (h - 1) = -h + 4 - 5 = -h - 1$.Since $x' = 1$,the reflected line is $x = 1$.The curve $C$ is the reflection of $y^2 = 4x$. The intersection points $A$ and $B$ of $C$ with $y = -5$ correspond to the intersection points of $y^2 = 4x$ with the line $x = 1$.Substituting $x = 1$ into $y^2 = 4x$,we get $y^2 = 4(1) = 4$,so $y = \pm 2$.The points are $(1, 2)$ and $(1, -2)$.The distance between these points is $|2 - (-2)| = 4$.

Let the curve $C$ be the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y=-5$,then the distance between $A$ and $B$ is

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