A) \[12{{x}^{2}}+4{{y}^{2}}=3\]
B) \[12{{x}^{2}}-4{{y}^{2}}=3\]
C) \[12{{x}^{2}}-4{{y}^{2}}+3=0\]
D) \[12{{x}^{2}}+4{{y}^{2}}+3=0\]
Correct Answer: B
Solution :
According to the given condition \[\sqrt{{{(x-1)}^{2}}+{{y}^{2}}}-\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=\pm 1\] On squaring both sides, we get \[2{{x}^{2}}+2{{y}^{2}}+1=2\sqrt{{{(x-1)}^{2}}+{{y}^{2}}}.\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}\] Again on squaring, we get\[12{{x}^{2}}-4{{y}^{2}}=3\].
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