Conic Sections Question 125
Question: The point on the parabola $ y^{2}=8x $ at which the normal is parallel to the line $ x-2y+5=0 $ is
Options:
A) $ (-1/2,\ 2) $
B) $ (1/2,\ -2) $
C) $ (2,\ -1/2) $
D) $ (-2,\ 1/2) $
Show Answer
Answer:
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} $
therefore $ k=-2 $
Substituting $ (h,k) $ and $ k=-2 $ , we get $ h=\frac{1}{2} $
Hence point is $ ( \frac{1}{2},-2 ) $ .
Trick: Here only point $ ( \frac{1}{2},-2 ) $ satisfies the parabola $ y^{2}=8x $ .
BETA
