A) \[{{y}^{2}}-2ax={{a}^{2}}\]
B) \[{{y}^{2}}-2ax+{{a}^{2}}=0\]
C) \[{{y}^{2}}+2ax+{{a}^{2}}=0\]
D) \[{{y}^{2}}+2ax={{a}^{2}}\]
Correct Answer: B
Solution :
Let the point be \[(h,\,\,k),\,\] So,\[{{(h-a)}^{2}}+{{(k-0)}^{2}}={{h}^{2}}\]\[\Rightarrow \,\,\,{{h}^{2}}+{{a}^{2}}-2ah+{{k}^{2}}={{h}^{2}}\] Hence locus is\[{{y}^{2}}-2ax+{{a}^{2}}=0\].
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