A) \[2\sqrt{3}y=12x+1\]
B) \[2\sqrt{3}y=-x-12\]
C) \[\sqrt{3}y=x+3\]
D) \[\sqrt{3y}=3x+1\]
Correct Answer: C
Solution :
Tangent to \[{{y}^{2}}=\,4x\] with slope m is \[y\,=\,mx\,+\frac{1}{m}\]\[{{m}^{2}}x-my+1=0\] is tangent to the circle \[{{x}^{2}}+{{y}^{2}}-6x=0\] |
\[\Rightarrow \] \[d=r\] |
\[\therefore \] \[\left| \frac{3{{m}^{2}}+1}{\sqrt{{{m}^{4}}+{{m}^{2}}}} \right|\,=\,3\] |
\[\Rightarrow \] \[\text{m}\,=\,\pm \,\frac{1}{\sqrt{3}}\] |
For \[m\,=\,\frac{1}{\sqrt{3}},\]is tangent to the circle \[y\,=\,\frac{x}{\sqrt{3}}+\sqrt{3}\] |
i.e., \[\sqrt{3y}\,=\,x+3.\] |
You need to login to perform this action.
You will be redirected in
3 sec