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  • The mirror image of the directrix of parabola     in the

The mirror image of the directrix of parabola   \mathrm {y^2=4(x-1)}  in the line mirror  \mathrm { x-2 y=2 }  is
 

Option: 1

\mathrm {3 x-4 y+4=0}


Option: 2

\mathrm {3 x+4 y+4=0}


Option: 3

\mathrm {4 x-3 y+4=0}


Option: 4

\mathrm {4 x+3 y-4=0}


Answers (1)

best_answer

Given parabola is  \mathrm {y^2=4(x-1)} 
\mathrm {\Rightarrow y^2=4 X=4(1) X( where \: \: x-1=X) }

\therefore  Directrix equation of given parabola is  \mathrm {x=0}
Any point on the line  \mathrm {x=0 (at \: \: y-axis)\: is\: (0, k)}
Now, mirror image of   \mathrm { (0, k) }  in the line   \mathrm { x-2 y=2 }  is given by
\begin{aligned} &\mathrm { \frac{x-0}{1}=\frac{y-k}{-2}=\frac{-2(0-2 k-2)}{5} }\\ & \mathrm {\text { or } \frac{x}{1}=\frac{y-k}{-2}=\frac{4 k+4}{5}} \\ & \mathrm {\quad x=\frac{4 k+4}{5} \text { or } k=\frac{5 x-4}{4} \quad(0, k)} \\ & \mathrm {\text { and } y-k=\frac{-8 k-8}{5}} \\ &\mathrm { \Rightarrow y=\frac{-(3 k+8)}{5}, k=\frac{5 y+8}{-3} \ldots\left(^{* *}\right)} \\ & \mathrm {\Rightarrow \frac{5 y+8}{-3}=\frac{(5 x-4)}{4} \quad\left(\text { From }\left(^*\right) \&(* *)\right) }\\ &\mathrm { \Rightarrow 4 y+3 x+4=0 \text { or } 3 x+4 y+4=0} \end{aligned}

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