A) \[\sqrt{3}y=3x+1\]
B) \[\sqrt{3}y=-(x+3)\]
C) \[\sqrt{3}y=x+3\]
D) \[\sqrt{3}y=-(3x+1)\]
Correct Answer: C
Solution :
Any tangent to \[{{y}^{2}}=4x\] is \[y=mx+\frac{1}{m}.\] It touches the circle, if \[3=\left| \frac{3m+\frac{1}{m}}{\sqrt{1+{{m}^{2}}}} \right|\]You need to login to perform this action.
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