A) \[{{x}^{2}}+4x-8y-12=0\]
B) \[{{x}^{2}}-4x+8y+12=0\]
C) \[{{x}^{2}}+8y=12\]
D) \[{{x}^{2}}-4x+12=0\]
Correct Answer: B
Solution :
\[VS=\sqrt{{{(2-2)}^{2}}+{{(-3+1)}^{2}}}=2\]. From \[{{(x-h)}^{2}}=-4a(y-k)\] Parabola is, \[{{(x-2)}^{2}}=-4.2(y+1)\] Þ \[{{(x-2)}^{2}}=-\,8(y+1)\] Þ \[{{x}^{2}}+4-4x=-8y-8\] Þ \[{{x}^{2}}-4x+8y+12=0.\]You need to login to perform this action.
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