A) \[4\]
B) \[-6\]
C) \[6\]
D) \[-4\]
Correct Answer: B
Solution :
Let the equation of the lines by \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\], then \[{{m}_{2}}=m_{1}^{2}\] ... (i) Also, \[{{m}_{1}}+{{m}_{2}}=\frac{2h}{b}\] \[{{m}_{1}}+m_{1}^{2}=\frac{-2h}{b}\] ... (ii) and \[{{m}_{1}}{{m}_{2}}=\frac{a}{b}\Rightarrow m_{1}^{3}=\frac{a}{b}\] ... (iii) Taking cube of both the sides of the Eq. (ii), we get \[m_{1}^{3}+m_{1}^{6}+m_{1}^{3}({{m}_{1}}+m_{1}^{2})=\frac{8{{h}^{3}}}{{{b}^{3}}}\] Again, from Eqs. (ii) and (iii), we get \[\left( \frac{a}{b} \right)+{{\left( \frac{a}{b} \right)}^{2}}+3\left( \frac{a}{b} \right)\left( \frac{-2h}{b} \right)=\frac{-8{{h}^{3}}}{{{b}^{3}}}\] \[\Rightarrow \] \[{{a}^{2}}b+a{{b}^{2}}+8{{h}^{2}}=6abh\] On dividing both sides by\[abh\], we get \[\frac{a+b}{h}+\frac{8{{h}^{2}}}{ab}=6\]
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