A) \[{{y}^{2}}-2ax+{{a}^{2}}=0\]
B) \[{{y}^{2}}+2ax+{{a}^{2}}=0\]
C) \[{{x}^{2}}-2ay+{{a}^{2}}=0\]
D) \[{{x}^{2}}+2ay+{{a}^{2}}=0\]
Correct Answer: A
Solution :
Accordingly,\[{{(h-a)}^{2}}+{{k}^{2}}={{h}^{2}}\] Þ \[-2ah+{{a}^{2}}+{{k}^{2}}=0\] Replace (h, k) by (x, y), then \[{{y}^{2}}-2ax+{{a}^{2}}=0\] is the required locus.You need to login to perform this action.
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