A) \[\frac{a+b}{h}+\frac{8{{h}^{2}}}{ab}+6=0\]
B) \[\frac{a-b}{h}+\frac{8{{h}^{2}}}{ab}+6=0\]
C) \[\frac{a+b}{h}+\frac{8{{h}^{2}}}{ab}=6\]
D) \[\frac{a-b}{h}+\frac{8h}{a{{b}^{2}}}=6\]
Correct Answer: C
Solution :
Given equation is \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] Let slope of both the lines represented by above equation are m and\[{{m}^{2}},\]then \[m+{{m}_{2}}=-\frac{2h}{b}\] ??(i) \[\left[ \because {{m}_{1}}+{{m}_{2}}=\frac{-2h}{b},{{m}_{1}}{{m}_{2}}=\frac{a}{b} \right]\] and \[m.{{m}^{2}}=\frac{a}{b}\] ?(ii) \[\Rightarrow \] \[m={{\left( \frac{a}{b} \right)}^{1/3}}\] Putting this value of\[m\]is Eq. (i), we get \[{{\left( \frac{a}{b} \right)}^{1/3}}+{{\left( \frac{a}{b} \right)}^{2/3}}=-\frac{2h}{b}\] ?.(iii) \[\Rightarrow \] \[{{\left[ {{\left( \frac{a}{b} \right)}^{1/3}}+{{\left( \frac{a}{b} \right)}^{2/3}} \right]}^{3}}={{\left( -\frac{2h}{b} \right)}^{3}}\] \[\Rightarrow \] \[\frac{a}{b}+{{\left( \frac{a}{b} \right)}^{2}}+3{{\left( \frac{a}{b} \right)}^{1/3}}.{{\left( \frac{a}{b} \right)}^{2/3}}\] \[\left\{ {{\left( \frac{a}{b} \right)}^{1/3}}+{{\left( \frac{a}{b} \right)}^{2/3}} \right\}=-\frac{8{{h}^{3}}}{{{b}^{3}}}\] \[\Rightarrow \] \[\frac{a}{b}+\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{3a}{b}\left( -\frac{2h}{b} \right)=-\frac{8{{h}^{3}}}{{{b}^{3}}}\] [Using Eq.(iii)] \[\Rightarrow \] \[\frac{ab+{{a}^{2}}}{{{b}^{2}}}+\frac{8{{h}^{3}}}{{{b}^{3}}}=\frac{6ah}{{{b}^{2}}}\] \[\Rightarrow \] \[\frac{a(a+b)}{{{b}^{2}}}+\frac{8{{h}^{3}}}{{{b}^{3}}}=\frac{6ah}{{{b}^{2}}}\] \[\Rightarrow \] \[\frac{a+b}{h}+\frac{8{{h}^{2}}}{ab}=6\] (On multiplying by\[\frac{{{b}^{2}}}{ah}\])You need to login to perform this action.
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