question_answer

Equation of mirror image of parabola\[{{y}^{2}}=4x\] in the line \[x-y+2=0\]is:

A) \[4\,(y-2)={{(x-2)}^{2}}\]        

B) \[4\,(y+2)={{(x+2)}^{2}}\]

C) \[4\,(y-2)={{(x+2)}^{2}}\]      

D) \[(y-2)={{(x+2)}^{2}}\]

Correct Answer: C

Solution :

\[\frac{x-{{t}^{2}}}{1}=\frac{y-2t}{-1}=-2\frac{({{t}^{2}}-2t+2)}{2}\]

\[\Rightarrow \]   \[x=2t-2\]          \[\Rightarrow \]   \[t=\frac{x+2}{2}\]

and       \[y={{t}^{2}}+2\]       \[\Rightarrow \]   \[y-2=\frac{{{(x+2)}^{2}}}{4}\]

\[\Rightarrow \]   \[4\,(y-2)={{(x+2)}^{2}}\]