Pair Of Straight Lines Question 113
Question: If the pair of straight lines given by $ Ax^{2}+2Hxy+By^{2}=0 $ , $ (H^{2}>AB) $ forms an equilateral triangle with line $ ax+by+c=0 $ , then $ (A+3B)(3A+B) $ is
[EAMCET 2003]
Options:
A) $ H^{2} $
B) $ -H^{2} $
C) $ 2H^{2} $
D) $ 4H^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
We know that the pair of lines $ (a^{2}-3b^{2})x^{2}+8abxy+(b^{2}-3a^{2})y^{2}=0 $ with the line $ ax+by+c=0 $ form an equilateral triangle.
Hence comparing with $ Ax^{2}+2Hxy+By^{2}=0 $ , then $ A=a^{2}-3b^{2},B=b^{2}-3a^{2},2H=8ab $ .
Now, $ (A+3B)(3A+B)=(-8a^{2})(-8b^{2}) $ = $ {{(8ab)}^{2}}={{(2H)}^{2}}=4H^{2} $
BETA
