Pair Of Straight Lines Question 106
Question: If the bisectors of the angles between the pairs of lines given by the equation $ ax^{2}+2hxy+by^{2}=0 $ and $ ax^{2}+2hxy+by^{2}+\lambda (x^{2}+y^{2})=0 $ be coincident, then $ \lambda = $
Options:
A) a
B) b
C) $ h $
D) Any real number
Show Answer
Answer:
Correct Answer: D
Solution:
Bisectors of $ ax^{2}+2hxy+by^{2}=0 $ are $ \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h} $ …..(i)
and of $ ax^{2}+2hxy+by^{2}+\lambda (x^{2}+y^{2})=0 $
i.e., $ (a+\lambda )x^{2}+2hxy+(b+\lambda )y^{2}=0 $
are $ \frac{x^{2}-y^{2}}{(a+\lambda )-(b+\lambda )}=\frac{xy}{h} $ …..(ii)
Which is the same equation as equation (i).
Hence for any $ \lambda $ belonging to real numbers, the lines will have same bisectors.
BETA
