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

Solution

Equation of ${\text{AA}}^{\prime }$ is
$\mathrm{y}+5=1\left(\mathrm{x}-\mathrm{\alpha }\right)$
$\Rightarrow \mathrm{y}=\mathrm{x}-\mathrm{\alpha }-5$
$\mathrm{y}-\mathrm{x}=-\mathrm{\alpha }-5$ ...(i)
Also
$\mathrm{x}+\mathrm{y}=-4$
$\mathrm{y}=\frac{-\mathrm{\alpha }-9}{2}$
$2\mathrm{x}=\mathrm{\alpha }+1$
$\Rightarrow \mathrm{x}=\frac{\mathrm{\alpha }+1}{2}$
$\mathrm{M}\equiv \left(\frac{\mathrm{\alpha }+1}{2},\frac{-\mathrm{\alpha }-9}{2}\right)$
${\mathrm{A}}^{\mathrm{\prime }}=\left(1,\mathrm{ }\mathrm{\alpha }-4\right)$
${\mathrm{A}}^{\mathrm{\prime }}$ lies on parabola
Now,
${\mathrm{\alpha }}^2+16+8\mathrm{\alpha }=4$
$\Rightarrow {\mathrm{\alpha }}^2+8\mathrm{\alpha }+12=0$
$\Rightarrow \mathrm{\alpha }=-2,\mathrm{ }-6$
AB = distance = 4 units

Hence, the correct answer is 4.