Conic Sections Question 41
Question: The directrix of the parabola $ x^{2}-4x-8y+12=0 $ is
[Karnataka CET 2003]
Options:
A) $ x=1 $
B) $ y=0 $
C) $ x=-1 $
D) $ y=-1 $
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Answer:
Correct Answer: D
Solution:
Equation of parabola is $ x^{2}-4x-8y+12=0 $
therefore $ x^{2}-4x+4=8y-8 $
therefore $ {{(x-2)}^{2}}=8(y-1) $
therefore $ X^{2}=8Y $
Comparing with $ X^{2}=4aY $ , we get $ a=2 $
Directrix is $ Y=-a $
therefore $ y-1=-2 $
therefore $ y=-1 $ .
BETA
