A) \[{{x}^{2}}+2x-2y-3=0\]
B) \[2{{x}^{2}}=3y\]
C) \[{{x}^{2}}-2x-y+3=0\]
D) None of these
Correct Answer: A
Solution :
\[{{(x+1)}^{2}}=4a(y+2)\] Passes through (3, 6) Þ \[16=4a.8\] Þ \[a=\frac{1}{2}\] Þ \[{{(x+1)}^{2}}=2(y+2)\] Þ \[{{x}^{2}}+2x-2y-3=0\].You need to login to perform this action.
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