A) \[\frac{{{a}^{2}}}{{{p}^{2}}}\]
B) \[\frac{{{b}^{2}}}{{{q}^{2}}}\]
C) \[\frac{{{c}^{2}}}{{{r}^{2}}}\]
D) none of these
Correct Answer: A
Solution :
Given\[\alpha ,\beta \] be the roots of \[a{{x}^{2}}+bx+c=0\]and\[\alpha +h,\beta +h\] are the roots of \[p{{x}^{2}}+qx+r=0\] and also\[{{D}_{1}},{{D}_{2}}\] discriminants of respectively equations. Let \[A=\alpha +h\]and \[B=\beta +h\] Then, \[A-B=(\alpha +h)-(\beta +h)=\alpha -\beta \] \[\Rightarrow \] \[{{(A-B)}^{2}}={{(\alpha -\beta )}^{2}}\] \[\Rightarrow \] \[{{(A+B)}^{2}}-4AB={{(\alpha +\beta )}^{2}}-4\,\alpha \beta \] \[\Rightarrow \] \[\frac{{{q}^{2}}}{{{p}^{2}}}-\frac{4r}{p}=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{4c}{a}\] \[\Rightarrow \] \[\frac{{{q}^{2}}-4rp}{{{p}^{2}}}=\frac{{{b}^{2}}-4ac}{{{a}^{2}}}\] \[\Rightarrow \] \[\frac{{{D}_{2}}}{{{p}^{2}}}=\frac{{{D}_{1}}}{{{a}^{2}}}\] \[(\because \,{{D}_{1}}={{b}^{2}}-4ac,{{D}_{2}}={{q}^{2}}-4rp)\] \[\Rightarrow \] \[\frac{{{D}_{1}}}{{{D}_{2}}}=\frac{{{a}^{2}}}{{{p}^{2}}}\]
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