Question

If $t₁, t₂$ and $t₃$ are distinct, the points $(t₁, 2at₁+at₁³), (t₂, 2at₂+at₂³), (t₃, 2at₃+at₃³)$ are collinear if

• A. $t_{1}t_{2}t_{3}=1$
• B. $t_{1}+t_{2}+t_{3}=t_{1}t_{2}t_{3}$
• C. $t_{1}+t_{2}+t_{3}=0$
• D. $t_{1}+t_{2}+t_{3}=-1$

Hint:

If $t1, t2$ and $t3$ are distinct, the points $(t1, 2at1+at1), (t2, 2at2+at2), (t3, 2at3+at3)$ are collinear if

Answer 1

The Correct Option is C

Step 1: Concept
Three points are collinear if the determinant of their coordinates is zero.
Step 2: Analysis
$\begin{vmatrix} t_{1} & 2at_{1}+at_{1}^{3} & 1 \\ t_{2} & 2at_{2}+at_{2}^{3} & 1 \\ t_{3} & 2at_{3}+at_{3}^{3} & 1 \end{vmatrix} = 0$. Factoring out '$a$', we get $a\begin{vmatrix} t_{1} & 2t_{1}+t_{1}^{3} & 1 \\ t_{2} & 2t_{2}+t_{2}^{3} & 1 \\ t_{3} & 2t_{3}+t_{3}^{3} & 1 \end{vmatrix} = 0$.
Step 3: Evaluation
Applying row operations $R_{2} \to R_{2}-R_{1}$ and $R_{3} \to R_{3}-R_{1}$ and simplifying , the determinant yields $(t_{2}-t_{1})(t_{3}-t_{1})(t_{3}-t_{2})(t_{1}+t_{2}+t_{3}) = 0$.
Step 4: Conclusion
Since the $t$ values are distinct ($t_{1} \ne t_{2} \ne t_{3}$), we must have $t_{1}+t_{2}+t_{3}=0$.
Final Answer: (c)