Conic Sections Question 195
Question: The equation of the common tangent to the curves $ y^{2}=8x $ and $ xy=-1 $ is
[IIT Screening 2002]
Options:
A) $ 3y=9x+2 $
B) $ y=2x+1 $
C) $ 2y=x+8 $
D) $ y=x+2 $
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Answer:
Correct Answer: D
Solution:
Any point on $ y^{2}=8x $ is $ (2t^{2},4t) $ where the tangent is $ yt=x+2t^{2}. $
Solving it with $ xy=-1, $
$ y(yt-2t^{2})=-1 $
or $ ty^{2}-2t^{2}y+1=0 $ . For common tangent, it should have equal roots. $ 4t^{4}-4t=0 $
therefore $ t=0,1 $ .
$ \therefore $ The common tangent is $ y=x+2 $ , (when $ t=0 $ , it is $ x=0 $ which can touch $ xy=-1 $ at infinity only).
BETA
