A) \[(6,\ -4\sqrt{3})\]
B) \[(6,\ 4\sqrt{3})\]
C) \[(-6,\ -4\sqrt{3})\]
D) \[(-6,\ 4\sqrt{3})\]
Correct Answer: A
Solution :
Normal at \[(h,k)\]to the parabola \[{{y}^{2}}=8x\] is \[y-k=-\frac{k}{4}(x-h)\] Gradient\[=\tan 60{}^\circ =\sqrt{3}=-\frac{k}{4}\]Þ \[k=-4\sqrt{3}\] and \[h=6\] Hence required point is \[(6,-4\sqrt{3})\].You need to login to perform this action.
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