A) \[(-1/2,\,2)\]
B) \[(1/2,\,-2)\]
C) \[(2,\,-1/2)\]
D) \[(-2,\,1/2)\]
Correct Answer: B
Solution :
Let point on the parabola be \[(h,k).\] Equation of the normal is, \[y-k=\frac{-k}{4}(x-h)\] or \[-kx-4y+kh+4k=0\] Since, normal is parallel to the line \[x-2y+5=0.\] \[\therefore \,\,\frac{-k}{4}=\frac{1}{2}\Rightarrow k=-2\] \[\because \,\,(h,k)\] lies on the parabola. \[\therefore \,\,{{k}^{2}}=8h\] \[\Rightarrow \,\,h=\frac{1}{2}\] Hence, point is \[\left( \frac{1}{2},-2 \right)\].
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