A) \[4(y+1)={{(x+1)}^{2}}\]
B) \[4(y-1)={{(x-1)}^{2}}\]
C) \[4(y+1)={{(x-1)}^{2}}\]
D) \[4(y-1)={{(x+1)}^{2}}\]
Correct Answer: D
Solution :
[d] Image of \[({{\lambda }^{2}},2\lambda )\] in the line mirror \[x-y+1=0\] is \[\frac{h-{{\lambda }^{2}}}{1}=\frac{k-2\lambda }{-1}=\frac{-2({{\lambda }^{2}}-2\lambda +1)}{2}\] \[\Rightarrow \,\,\,\,\,\,h+1=2\lambda \] ....(1) and \[k={{\lambda }^{2}}+1\] ....(2) Putting value of \[\lambda \] from (1) into (2), we get \[k-1={{\left( \frac{h+1}{2} \right)}^{2}}\] or \[4(y-1)={{(x+1)}^{2}},\] which is the required locus.You need to login to perform this action.
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