Question

The mirror image of any point on the directrix of $y^2 = 4(x + 1)$ lies on

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

Answer 1

Answer: (B)

$3x - 4y +16 = 0$

Answer 2

Directrix of $y^{2}= 4\left(x+1\right)$ is $x=-2$ Any point on it is $\left(-2, k\right)$ Now mirror image (x, y) of (- 2, k) in the line x + 2y = 3 is given by $Also y = k +4 -\frac{8k}{5}$ y = k +4 - $\frac{8k}{5}$ Hence y = 4 - $\frac{3k}{5}=4+\frac{3}{5}\left(\frac {5x}{4}\right)=\frac {16+3x}{4}$ $\Rightarrow\,\,$3x -4y + 16 = 0 is the reqd. mirror image.