Pair Of Straight Lines Question 119
Question: If the bisectors of the lines $ x^{2}-2pxy-y^{2}=0 $ be $ x^{2}-2qxy-y^{2}=0, $ then
[MP PET 1993; DCE 1999; RPET 2003; AIEEE 2003; Kerala (Engg.) 2005]
Options:
A) $ pq+1=0 $
B) $ pq-1=0 $
C) $ p+q=0 $
D) $ p-q=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Bisector of the angle between the lines $ x^{2}-2pxy-y^{2}=0 $ is $ \frac{x^{2}-y^{2}}{xy}=\frac{1-(-1)}{-p} $
$ \Rightarrow px^{2}+2xy-py^{2}=0 $
But it is represented by $ x^{2}-2qxy-y^{2}=0 $ .
Therefore $ \frac{p}{1}=\frac{2}{-2q}\Rightarrow pq=-1 $ .
BETA
