Conic Sections Question 53
Question: The point of contact of the tangent $ 18x-6y+1=0 $ to the parabola $ y^{2}=2x $ is
Options:
A) $ ( \frac{-1}{18},\ \frac{-1}{3} ) $
B) $ ( \frac{-1}{18},\ \frac{1}{3} ) $
C) $ ( \frac{1}{18},\ \frac{-1}{3} ) $
D) $ ( \frac{1}{18},\ \frac{1}{3} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Let point of contact be (h, k), then tangent at this point is $ ky=x+h $ . $ x-ky+h=0\equiv 18x-6y+1=0 $
or $ \frac{1}{18}=\frac{k}{6}=\frac{h}{1} $ or $ k=\frac{1}{3} $ , $ h=\frac{1}{18} $ .
BETA
