A) \[y=x+1\]
B) \[y=x+2\]
C) \[y=x-2\]
D) \[y=-x+2\]
Correct Answer: B
Solution :
Given, parabola\[{{y}^{2}}=8x\] where \[a=2\] Equation of tangent to the parabola is \[y=mx+\frac{a}{m}\] \[\Rightarrow \] \[mx-y=\frac{2}{m}=0\] ?.(i) If it touches the parabola\[{{x}^{2}}+{{y}^{2}}=2,\]then length of perpendicular drawn from its centre (0, 0) is equal to the radius. \[\therefore \] \[\frac{0-0+\frac{2}{m}}{\sqrt{{{m}^{2}}+1}}=\sqrt{2}\] \[\Rightarrow \] \[\frac{2}{m}=\sqrt{2}\sqrt{{{m}^{2}}+1}\] \[\Rightarrow \] \[\frac{4}{{{m}^{2}}}=2({{m}^{2}}+1)\] \[\Rightarrow \] \[{{m}^{2}}({{m}^{2}}+1)=2\] \[\Rightarrow \] \[{{m}^{4}}+{{m}^{2}}-2=0\] \[\Rightarrow \] \[({{m}^{2}}+2)({{m}^{2}}-1)=0\] \[\Rightarrow \] \[{{m}^{2}}\ne -2,{{m}^{2}}=1\] \[\Rightarrow \] \[m=\pm 1\] On putting the value of \[m\]in Eq. (i), \[\pm \text{ }x-y\pm 2=0\] Hence, equation of common tangent is \[y=\pm (x+2)\]
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