A) \[y=x+1\]
B) \[y=x+2\]
C) \[y=x-2\]
D) \[y=-x+2\]
Correct Answer: B
Solution :
\[{{y}^{2}}=8x,\] \[\therefore 4a=8\] Þ \[a=2\] Any tangent of parabola is, \[y=mx+\frac{a}{m}\] or \[mx-y+\frac{2}{m}=0\] If it is a tangent to the circle \[{{x}^{2}}+{{y}^{2}}=2,\] then perpendicular from centre \[(0,0)\] is equal to radius \[\sqrt{2}\]. \[\therefore \frac{2/m}{\sqrt{{{m}^{2}}+1}}=\sqrt{2}\] or \[\frac{4}{{{m}^{2}}}=2({{m}^{2}}+1)\] Þ \[{{m}^{4}}+{{m}^{2}}-2=0\] Þ \[({{m}^{2}}+2)({{m}^{2}}-1)=0\] or \[m=\pm 1\] Hence the common tangent are \[y=\pm (x+2)\] \[\therefore y=x+2\].
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