• # question_answer Let $({{x}_{i}},{{y}_{i}})i=1,2,3,4$ are the integral solutions of equation $2{{x}^{2}}{{y}^{2}}+{{y}^{2}}-6{{x}^{2}}-12=0.$ Then area of quadrilateral whose vertices are $({{x}_{i}},{{y}_{i}})$is 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 = 16

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