A) \[{{t}_{1}}{{t}_{2}}=-1\]
B) \[{{t}_{1}}{{t}_{2}}=1\]
C) \[2{{t}_{1}}{{t}_{2}}={{t}_{1}}+{{t}_{2}}\]
D) \[{{t}_{1}}+{{t}_{2}}=a\]
Correct Answer: C
Solution :
If there points \[A(at_{1}^{2},2a{{t}_{1}}),\,\,\,B(at_{2}^{2},\,2a{{t}_{2}})\] and \[C(0,a),\] collinear, if \[\left| \begin{matrix} at_{1}^{2} & 2a{{t}_{1}} & 1 \\ at_{2}^{2} & 2a{{t}_{2}} & 1 \\ 0 & a & 1 \\ \end{matrix} \right|=0\] Use operation; \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\left| \begin{matrix} at_{1}^{2} & 2a{{t}_{1}} & 1 \\ a(t_{2}^{2}-t_{1}^{2}) & 2a({{t}_{2}}-{{t}_{1}}) & 0 \\ -at_{1}^{2} & a-2a{{t}_{1}} & 0 \\ \end{matrix} \right|=0\] Expand with respect to \[{{C}_{3}}\] \[a({{t}_{2}}-{{t}_{1}})\,({{t}_{2}}+{{t}_{1}})\,(a-2a{{t}_{1}})\] \[+2{{a}^{2}}t_{1}^{2}({{t}_{2}}-{{t}_{1}})=0\] \[\Rightarrow \] \[a({{t}_{2}}-{{t}_{1}})\{({{t}_{1}}+{{t}_{2}})\,(a-2a{{t}_{1}})+2at_{1}^{2}\}=0\] \[\Rightarrow \] \[a({{t}_{2}}-{{t}_{1}})\,\{a{{t}_{1}}+a{{t}_{2}}-2at_{1}^{2}\] \[-2a{{t}_{1}}{{t}_{2}}+2at_{1}^{2}\}=0\] \[\Rightarrow \] \[a({{t}_{2}}-{{t}_{1}})\,(a{{t}_{1}}+a{{t}_{2}}-2a{{t}_{1}}{{t}_{2}})=0\] \[\Rightarrow \] \[{{a}^{2}}({{t}_{2}}-{{t}_{1}})\,({{t}_{1}}+{{t}_{2}}-2{{t}_{1}}+{{t}_{2}})=0\] \[\Rightarrow \] \[{{t}_{2}}-{{t}_{1}}=0\] or \[{{t}_{1}}+{{t}_{2}}-2{{t}_{1}}{{t}_{2}}=0\] \[\Rightarrow \] \[{{t}_{1}}={{t}_{2}}\] or \[{{t}_{1}}={{t}_{2}}\]
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