Pair Of Straight Lines Question 117
Question: If one of the line represented by the equation $ ax^{2}+2hxy+by^{2}=0 $ is coincident with one of the line represented by $ {a}‘x^{2}+2{h}‘xy+{b}‘y^{2}=0 $ , then
Options:
A) $ {{(a{b}’-{a}‘b)}^{2}}=4(a{h}’-{a}‘h)(h{b}’-{h}‘b) $
B) $ {{(a{b}’+{a}‘b)}^{2}}=4(a{h}’-{a}‘h)(h{b}’-{h}‘b) $
C) $ {{(a{b}’-{a}‘b)}^{2}}=(a{h}’-{a}‘h)(h{b}’-{h}‘b) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The equation of given lines are $ ax^{2}+2hxy+by^{2}=0 $ …..(i)
$ a’x^{2}+2h’xy+b’y^{2}=0 $ …..(ii)
Let common line to both is $ y=mx $ , then it will satisfy both the above equations.
Hence, $ a+2mh+bm^{2}=0 $ …..(iii)
and $ a’+2mh’+b’m^{2}=0 $ …..(iv)
Now eliminating ’m’ from the equation (iii) and (iv),
we get $ \frac{m^{2}}{2ha’-2h’a}=\frac{-m}{ba’-b’a}=\frac{1}{2bh’-2b’h} $
$ \Rightarrow m^{2}=\frac{ha’-h’a}{bh’-b’h} $ …..(v) and $ m^{2}=\frac{{{(ab’-ba’)}^{2}}}{4{{(bh’-b’h)}^{2}}} $ …..(vi)
From (v) and (vi), we get the required condition.
BETA
