A) \[y-3=0\]
B) \[y+3=0\]
C) \[3y+1=0\]
D) \[3y-1=0\]
Correct Answer: D
Solution :
On solving both the equations, we get \[\frac{8y}{3}+{{y}^{2}}=1\] \[\Rightarrow \] \[3{{y}^{2}}+8y-3=0\] \[\Rightarrow \] \[(3y-1)\,(y+3)=0\] \[\Rightarrow \] \[y=-3,\,\,\frac{1}{3},\] here \[y\ne -3\] \[\therefore \] At \[y=\frac{1}{3}\] \[\Rightarrow \] \[x=\pm \,2\sqrt{\frac{2}{3}}\] \[\therefore \] Point of intersection is \[\left( 2\sqrt{\frac{2}{3}},\,\frac{1}{3} \right)\] and \[\left( -2\sqrt{\frac{2}{3}},\,\frac{1}{3} \right)\] From option (d); \[3y-1=0\] is the required equation which satisfied the intersection points.You need to login to perform this action.
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