Conic Sections Question 50
Question: If $ x^{2}+6x+20y-51=0 $ , then axis of parabola is
[Orissa JEE 2004]
Options:
A) $ x+3=0 $
B) $ x-3=0 $
C) $ x=1 $
D) $ x+1=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation of parabola is $ x^{2}+6x+20y-51=0 $
$ \Rightarrow $ $ x^{2}+6x=-20y+51 $
$ \Rightarrow $ $ {{(x+3)}^{2}}=-20y+60\Rightarrow {{(x+3)}^{2}}=-20(y-3) $
$ \Rightarrow $ $ {{(x+3)}^{2}}=-4.5(y-3) $
$ \therefore $ Axis of parabola is $ x+3=0 $ .
BETA
