Question

# What is the angle between the straight lines $$\left( { m }^{ 2 }-mn \right) y=\left( mn+{ n }^{ 2 } \right) x+{ n }^{ 3 }$$ and $$\left( mn+{ m }^{ 2 } \right) y=\left( mn-{ n }^{ 2 } \right) x+{ m }^{ 3 }$$, where $$m> n$$?

A
tan1(2mnm2+n2)
B
tan1(4m2n2m4n4)
C
tan1(4m2n2m4+n4)
D
45o

Solution

## The correct option is B $$\tan ^{ -1 }{ \left( \cfrac { 4{ m }^{ 2 }{ n }^{ 2 } }{ { m }^{ 4 }-{ n }^{ 4 } } \right) }$$Given $$(m^{2} - mn)y = (mn + n^{2})x + n^{3}$$$$(mn + m^{2})y = (mn - n^{2})x + m^{3}$$Slope of line $$(1) = \dfrac {mn + n^{2}}{m^{2} -mn}$$Slope of line $$(2) \rightarrow \dfrac {mn - n^{2}}{m^{2} + mn}$$By comparing with $$y = mx + c$$$$\tan \theta = \pm \left (\dfrac {m_{1} - m_{2}}{1 + m_{1}m_{2}}\right ), m_{1}$$ and $$m_{2}$$ are slope$$\therefore \tan \theta = \dfrac {\left (\dfrac {mn + n^{2}}{m^{2} -mn} - \dfrac {mn - n^{2}}{m^{2} + mn}\right )}{\dfrac {1 + (mn + n^{2}(mn - n^{2})}{(m^{2} - mn)(m^{2} + mn)}}$$$$\tan \theta = \dfrac {(mn + n^{2}) (m^{2} + mn) - (mn - n^{2})(m^{2} - mn)}{m^{4} - (mn)^{2} + (mn)^{2} - n^{4}}$$$$\tan \theta = \dfrac {m^{3}n + (mn)^{2} + (mn)^{2} + mn^{3} - m^{3}n + (m^{2}n^{2}) + (mn)^{2} - mn^{3}}{m^{4} - n^{4}}$$$$\tan \theta = \dfrac {4m^{2}n^{2}}{m^{4}- n^{4}}$$$$\therefore \theta = \tan^{-1} \left (\dfrac {4m^{2} n^{2}}{m^{4} - n^{4}}\right )$$.Mathematics

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