A) \[x+4y-2=0\]
B) \[x+2y=0\]
C) \[x+y+1=0\]
D) \[x-y+3=0\]
Correct Answer: C
Solution :
Let \[(2t,{{t}^{2}})\]be any point on the parabola. Centre of circle \[=(-g,-f)=(-3,0)\] For the distance between point P and centre of circle to be minimum, line drawn from the centre of circle to point P must be normal to the parabola at P. Slope of line joining centre of circle to point P \[P=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{{{t}^{2}}-0}{2t+3}\] Slope of tangent to parabola at\[P=\frac{dy}{dx}=\frac{x}{2}=t\] Slope of normal \[=\frac{-1}{t}\] Therefore, \[\frac{{{t}^{2}}-0}{2t+3}=\frac{-1}{t}\] \[\Rightarrow \]\[{{t}^{3}}+2t+3=0\] \[\Rightarrow \]\[(t+1)({{t}^{2}}-t+1)=0\] Real roots of above equation is \[t=-1\] Coordinate of \[P=(2t,{{t}^{2}})=(-2.1)\] Slope of tangent of parabola at \[P=\frac{dy}{dx}=\frac{x}{2}=t=-1\] Therefore, equation of tangent is \[(y-1)=(-1)(x+2)\] \[\Rightarrow \]\[x+y+1=0\] Hence, answer is option C.
You need to login to perform this action.
You will be redirected in
3 sec
