A) \[(3,\,\,2)\]
B) \[(5,\,\,6)\]
C) \[(4,\,\,-1)\]
D) \[(2,\,\,-3)\]
Correct Answer: B
Solution :
Let\[({{x}_{1}},\,\,{{y}_{1}})\]be the mid-point of the line joining the common points of the given line and the given parabola. Then, the equation of line is \[y{{y}_{1}}-4(x+{{x}_{1}})=y_{1}^{2}-8{{x}_{1}}\] \[\Rightarrow \] \[4x-y{{y}_{1}}+y_{1}^{2}-4{{x}_{1}}=0\] This line and\[2x-3y+8=0\]represents the same line. \[\therefore \] \[\frac{4}{2}=\frac{-{{y}_{1}}}{-3}=\frac{y_{1}^{2}-4{{x}_{1}}}{8}\] \[\Rightarrow \] \[{{y}_{1}}=6,\,\,y_{1}^{2}-4{{x}_{1}}=16\] \[\Rightarrow \] \[{{y}_{1}}=6,\,\,{{x}_{1}}=\frac{36-16}{4}\] \[\Rightarrow \] \[{{y}_{1}}=6,\,\,{{x}_{1}}=5\] \[\therefore \]The required point is (5, 6).
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