A) 4
B) 8
C) 12
D) 16
Correct Answer: D
Solution :
\[2{{x}^{2}}{{y}^{2}}+{{y}^{2}}-6{{x}^{2}}-12=0\Rightarrow (2{{x}^{2}}+1)({{y}^{2}}-3)=9\]\[\Rightarrow \]\[2{{x}^{2}}+1=\frac{9}{{{y}^{2}}-3}\Rightarrow y=\pm 2\And x=\pm 2\] \[\therefore \]vertices are \[(2,2),(2,-2),(-2,2)\And (-2,-2)\] Area = 16You need to login to perform this action.
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