Question

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___

Answer 1

Let $P(t^2,2t)$ be a point on the curve $y^2=4x$, whose image is $Q(x,y)$ on $x+y+4=0$, then
$\frac{x-t^2}{1}=\frac{y-2t}{1}=\frac{-2(t^2+2t+4)}{1^2+1^2}$
$\Rightarrow x=-2t-4$
and $y=-t^2-4$

Now the straight line $y=-5$ meets the mirror image.
$\therefore -t^2-4=-5\Rightarrow t^2=1\Rightarrow t=\pm 1$
Thus, points of intersection of $A$ and $B$ are (-6,-5) and (-2, -5).
$\therefore$ Distance, $AB=\sqrt{(-2+6{)}^2+(-5+5{)}^2}=4$