A) \[{{(a+b)}^{2}}\]
B) \[{{(a-b)}^{2}}\]
C) \[{{h}^{2}}+ab\]
D) \[{{h}^{2}}-ab\]
Correct Answer: D
Solution :
Given that \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] ….(i) Which is homogeneous equations representing pair of straight line each of which passing through the origin. Given one slope of line =m, Let another slope of line \[={{m}_{1}}\] Then, the lines are \[y=mx\] and \[y={{m}_{1}}x\] Now, \[(mx-y)({{m}_{1}}x-y)\] \[\Rightarrow \] \[m{{m}_{1}}{{x}^{2}}-{{m}_{1}}xy-mxy+{{y}^{2}}\] \[\Rightarrow \] \[m{{m}_{1}}.{{x}^{2}}-(m+{{m}_{1}})y.x+{{y}^{2}}\] …..(ii) On comparing Eqs. (i) and (ii), \[m+{{m}_{1}}=-\frac{2h}{b}\] …..(iii) \[m{{m}_{1}}=\frac{a}{b}\] ……(iv) From Eqs. (iii) and (iv), \[{{m}_{1}}=\left( -\frac{2h}{b}-m \right)\] \[\Rightarrow \] \[m\left( \frac{-2h}{b}-m \right)=\frac{a}{b}\] \[\Rightarrow \] \[-\frac{m}{b}(2h+mb)=\frac{a}{b}\] \[\Rightarrow \] \[-2mh-{{m}^{2}}b=a\] \[\Rightarrow \] \[-2mhb-{{m}^{2}}{{b}^{2}}=a\] \[\Rightarrow \] \[{{h}^{2}}+2mbh+{{m}^{2}}{{b}^{2}}=-ab+{{h}^{2}}\] \[\Rightarrow \] \[{{(h+mb)}^{2}}={{h}^{2}}-ab\]
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