Pair Of Straight Lines Question 133
Question: If the pair of lines $ ax^{2}+2(a+b)xy+by^{2}=0 $ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then
[AIEEE 2005]
Options:
A) $ 3a^{2}+10ab+3b^{2}=0 $
B) $ 3a^{2}+2ab+3b^{2}=0 $
C) $ 3a^{2}-10ab+3b^{2}=0 $
D) $ 3a^{2}-2ab+3b^{2}=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Angle between the given lines is $ \tan \frac{\pi }{4}=\frac{2\sqrt{{{(a-b)}^{2}}-ab}}{a+b} $
Therefore $ \frac{2\sqrt{{{(a+b)}^{2}}-ab}}{a+b}=1 $
Therefore $ 3a^{2}+2ab+3b^{2}=0 $
BETA
