Conic Sections Question 6
Question: The axis of the parabola $ 9y^{2}-16x-12y-57=0 $ is
[MNR 1995]
Options:
A) $ 3y=2 $
B) $ x+3y=3 $
C) $ 2x=3 $
D) $ y=3 $
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ 9y^{2}-16x-12y-57=0 $
therefore $ {{( y-\frac{2}{3} )}^{2}}=\frac{16}{9}( x+\frac{61}{16} ) $
Put $ y-\frac{2}{3}=Y $ and $ x+\frac{61}{16}=X $
therefore $ Y^{2}=4( \frac{4}{9} )X $
Axis of this parabola is $ Y=0 $
therefore $ y-\frac{2}{3}=0 $
therefore $ 3y=2 $ .
BETA
