Question

The mirror image of $y^2=4x$ about the line $x-y+1=0$ is

• A. $(x-1{)}^2=4(y+1)$
• B. $(x+1{)}^2=4(y+1)$
• C. $(x+1{)}^2=4(y-1)$
• D. $(x-1{)}^2=4(y-1)$

Answer 1

The correct option is C $(x+1{)}^2=4(y-1)$

Any point on $y^2=4x$ is $P(t^2,2t).$
Let the image of $P$ about $x-y+1=0$ be $(h,k).$
$\therefore \frac{h-t^2}{1}=\frac{k-2t}{-1}=\frac{-2(t^2-2t+1)}{1^2+(-1{)}^2}$
$\Rightarrow h-t^2=-k+2t=-(t^2-2t+1)$
$\Rightarrow h=2t-1$ and $k=t^2+1$
Eliminating $t,$ we get
$k-1={\left(\frac{h+1}{2}\right)}^2$
Hence, locus of $P$ is $(x+1{)}^2=4(y-1)$