Conic Sections Question 127
Question: The point on the parabola $ y^{2}=8x $ at which the normal is inclined at 60o to the x-axis has the co-ordinates
[MP PET 1993]
Options:
A) $ (6,\ -4\sqrt{3}) $
B) $ (6,\ 4\sqrt{3}) $
C) $ (-6,\ -4\sqrt{3}) $
D) $ (-6,\ 4\sqrt{3}) $
Show Answer
Answer:
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} $
therefore $ k=-4\sqrt{3} $ and $ h=6 $
Hence required point is $ (6,-4\sqrt{3}) $ .
BETA
