Conic Sections Question 98
Question: The line $ x-y+2=0 $ touches the parabola $ y^{2}=8x $ at the point
[Roorkee 1998]
Options:
A) $ (2,\ -4) $
B) $ (1,\ 2\sqrt{2}) $
C) $ (4,\ -4\sqrt{2}) $
D) (2, 4)
Show Answer
Answer:
Correct Answer: D
Solution:
Intersection of $ x-y+2=0 $ with $ y^{2}=8x $ is given by $ {{(x+2)}^{2}}=8x $
therefore $ x^{2}+4-4x=0 $
therefore $ {{(x-2)}^{2}}=0 $
$ \therefore x=2 $ and $ y=x+2=4 $ .
BETA
