A) \[2x\pm y+1=0\]
B) \[2x\pm y-1=0\]
C) \[x\pm 2y+1=0\]
D) \[x\pm 2y-1=0\]
Correct Answer: A
Solution :
Tangent to \[{{y}^{2}}=8x\]Þ \[y=mx+\frac{2}{m}\] Tangent to \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{3}=1\]Þ\[y=mx\pm \sqrt{{{m}^{2}}-3}\] On comparing, we get m = ±2 or tangent asYou need to login to perform this action.
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