Pair Of Straight Lines Question 144
Question: If the slope of one of the lines represented by $ ax^{2}+2hxy+by^{2}=0 $ be the square of the other, then
Options:
A) $ a^{2}b+ab^{2}-6abh+8h^{3}=0 $
B) $ a^{2}b+ab^{2}+6abh+8h^{3}=0 $
C) $ a^{2}b+ab^{2}-3abh+8h^{3}=0 $
D) $ a^{2}b+ab^{2}-6abh-8h^{3}=0 $
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Answer:
Correct Answer: A
Solution:
Here, $ m_1=m_2^{2}\Rightarrow m_2^{2}+m_2=\frac{-2h}{b} $ ….. (i) and $ m_2^{2}m_2=\frac{a}{b}\Rightarrow m_2={{( \frac{a}{b} )}^{1/3}} $ …..(ii)
Putting this value of $ m_2 $ in (i), we get $ {{{ {{( \frac{a}{b} )}^{1/3}} }}^{2}}+{{( \frac{a}{b} )}^{1/3}}=\frac{-2h}{b} $
On cubing both sides, we get $ {{( \frac{a}{b} )}^{2}}+\frac{a}{b}+3{{( \frac{a}{b} )}^{2/3}}.{{( \frac{a}{b} )}^{1/3}}.{ {{( \frac{a}{b} )}^{2/3}}+{{( \frac{a}{b} )}^{1/3}} }=\frac{-8h^{3}}{b^{3}} $
Therefore $ {{( \frac{a}{b} )}^{2}}+\frac{a}{b}-\frac{6ah}{b^{2}}=\frac{-8h^{3}}{b^{3}} $
$ { \because {{( \frac{a}{b} )}^{2/3}}+{{( \frac{a}{b} )}^{1/3}}=\frac{-2h}{b} } $
Therefore $ ab(a+b)-6abh+8h^{3}=0 $ .
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