A) \[y=\sqrt{1+4x},x\ge 0\]
B) \[x=\sqrt{1+4y},y\ge 0\]
C) \[x=\sqrt{1+2y},y\ge 0\]
D) \[y=\sqrt{1+2x},x\ge 0\]
Correct Answer: D
Solution :
\[\sqrt{{{h}^{2}}+{{k}^{2}}}=|h|+1\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}={{x}^{2}}+1+2x\] \[\Rightarrow {{y}^{2}}=1+2x\] \[\Rightarrow y=\sqrt{1+2x};x\ge 0.\]You need to login to perform this action.
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