Question

The mirror image of the directrix of the parabola $y^2=4(x+1)$ in the line mirror $x+2y=3$ is

• A. $x=-2$
• B. $4x-3y=16$
• C. $3x-4y+16=0$
• D. $noneofthese$

Answer 1

The correct option is C $3x-4y+16=0$

Directrix of $y^2=4(x+1)$ is $x=-2$

Any point on $x=-2$ is $(-2,k)$

Now, mirror image $(x,y)$ of $(-2,k)$ in the line $x+2y=3$ is given by

$\frac{x+2}{1}=\frac{y-k}{2}=-2\left(\frac{-2+2k-3}{5}\right)$

$\Rightarrow x=\frac{10-4k}{5}-2\Rightarrow x=-\frac{4k}{5}$ ...(1)

Also $y=\frac{20-3k}{5}$ ....(2)

From (1) and (2)

$y=4+\frac{3}{5}\left(\frac{5x}{4}\right)\Rightarrow 4y=16+3x$

$\Rightarrow 3x-4y+16=0$

is the equation of the mirror image of the directrix.