A) \[h=2ab\,\,(a+b)\]
B) \[h=ab\,\,(a+b)\]
C) \[3h=2ab\,\,(a+b)\]
D) \[2h=ab\,\,(a+b)\]
Correct Answer: D
Solution :
| \[{{a}^{3}}{{x}^{2}}-2hxy+{{b}^{3}}{{y}^{2}}=0\] |
| Let the slop of lines be \[{{m}_{1}}\]and \[{{m}_{2}}.\] |
| Then, |
| \[{{m}_{1}}+{{m}_{2}}=\frac{2h}{{{b}^{3}}},\]\[{{m}_{1}}{{m}_{2}}=\frac{{{a}^{3}}}{{{b}^{3}}}\] |
| Given, \[m_{2}^{2}={{m}_{1}}\]\[\Rightarrow \]\[m_{2}^{3}=\frac{{{a}^{3}}}{{{b}^{3}}}\] |
| \[\Rightarrow \] \[{{m}_{2}}=\frac{a}{b}\] |
| Also \[m_{2}^{2}+{{m}_{2}}=\frac{2h}{{{b}^{3}}}\] |
| \[\Rightarrow \] \[\frac{2h}{{{b}^{3}}}=\frac{a}{b}+\frac{{{a}^{2}}}{{{b}^{2}}}\]\[\Rightarrow \]\[ab+{{b}^{2}}=\frac{2h}{b}\] |
| \[\Rightarrow \] \[2h={{a}^{2}}b+a{{b}^{2}}=ab\,\,(a+b)\] |
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