A) \[y=-x+2\]
B) \[y=x-2\]
C) \[y=x+2\]
D) None of these
Correct Answer: C
Solution :
[c] Any tangent to parabola \[{{y}^{2}}=8x\]is y \[=mx+\frac{2}{m}\] ? (i) It touches the circle \[{{x}^{2}}+{{y}^{2}}-12x+4=0\], if the length of perpendicular from the centre (6, 0) is equal to radius \[\sqrt{32.}\] \[\therefore \frac{6m+\frac{2}{m}}{\sqrt{{{m}^{2}}+1}}=\pm \sqrt{32}\Rightarrow {{\left( 3m+\frac{1}{m} \right)}^{2}}=8({{m}^{2}}+1)\] \[\Rightarrow {{(3{{m}^{2}}+1)}^{2}}=8({{m}^{4}}+{{m}^{2}})\] \[\Rightarrow {{m}^{4}}-2{{m}^{2}}+1=0\Rightarrow m=\pm 1\] Hence, the required tangents are \[y=x+2\] and \[y=-x-2.\]
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