Solveeit Logo

Question

Mathematics Question on Determinants

The three distinct points A(at12,2at1),B(at22,2at2)A(at_{1}^{2},\,2a{{t}_{1}}),\,\,B(at_{2}^{2},\,\,2a{{t}_{2}}) and C(0,a)C(0,\,a) (where aa is a real number) are collinear, if

A

t1t2=1{{t}_{1}}{{t}_{2}}=-1

B

t1t2=1{{t}_{1}}{{t}_{2}}=1

C

2t1t2=t1+t22{{t}_{1}}{{t}_{2}}={{t}_{1}}+{{t}_{2}}

D

t1+t2=a{{t}_{1}}+{{t}_{2}}=a

Answer

2t1t2=t1+t22{{t}_{1}}{{t}_{2}}={{t}_{1}}+{{t}_{2}}

Explanation

Solution

If there points A(at12,2at1),B(at22,2at2)A(at_{1}^{2},2a{{t}_{1}}),\,\,\,B(at_{2}^{2},\,2a{{t}_{2}}) and C(0,a),C(0,a), collinear, if
at122at11 at222at21 0a1 =0\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; R2R2R1,R3R3R1{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} at122at11 a(t22t12)2a(t2t1)0 at12a2at10 =0\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 C3{{C}_{3}}
a(t2t1)(t2+t1)(a2at1)a({{t}_{2}}-{{t}_{1}})\,({{t}_{2}}+{{t}_{1}})\,(a-2a{{t}_{1}})
+2a2t12(t2t1)=0+2{{a}^{2}}t_{1}^{2}({{t}_{2}}-{{t}_{1}})=0
\Rightarrow a(t2t1)(t1+t2)(a2at1)+2at12=0a({{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(t2t1)(at1+at22at1t2)=0a({{t}_{2}}-{{t}_{1}})\,(a{{t}_{1}}+a{{t}_{2}}-2a{{t}_{1}}{{t}_{2}})=0
\Rightarrow a2(t2t1)(t1+t22t1+t2)=0{{a}^{2}}({{t}_{2}}-{{t}_{1}})\,({{t}_{1}}+{{t}_{2}}-2{{t}_{1}}+{{t}_{2}})=0
\Rightarrow t2t1=0{{t}_{2}}-{{t}_{1}}=0
or t1+t22t1t2=0{{t}_{1}}+{{t}_{2}}-2{{t}_{1}}{{t}_{2}}=0
\Rightarrow t1=t2{{t}_{1}}={{t}_{2}}
or t1=t2{{t}_{1}}={{t}_{2}}