A) \[(-1/2,\ 2)\]
B) \[(1/2,\ -2)\]
C) \[(2,\ -1/2)\]
D) \[(-2,\ 1/2)\]
Correct Answer: B
Solution :
Let point be \[(h,k).\] Normal is \[y-k=\frac{-k}{4}(x-h)\] or \[-kx-4y+kh+4k=0\] Gradient \[=-\frac{k}{4}=\frac{1}{2}\]Þ \[k=-2\] Substituting \[(h,k)\] and \[k=-2\], we get \[h=\frac{1}{2}\] Hence point is \[\left( \frac{1}{2},-2 \right)\]. Trick: Here only point \[\left( \frac{1}{2},-2 \right)\] satisfies the parabola\[{{y}^{2}}=8x\].
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