Question

Sol. [Key Concept: Apply the section formula at $C$ (on the axis) dividing $AB$ in $1:3$ with $A(t_{1}),B(t_{2})$; equate coordinates to relate $t_{1},t_{2}$.]

$$\begin{matrix} & \ \begin{matrix} t_{1}t_{2} & \ = - \frac{c}{a} \Rightarrow c = - at_{1}t_{2} \\ \frac{AC^{2}}{CB^{2}} & \ = \frac{\left( at_{1}^{2} + at_{1}t_{2} \right)^{2} + 4a^{2}t_{1}^{2}}{\left( at_{2}^{2} + at_{1}t_{2} \right)^{2} + 4a^{2}t_{2}^{2}} = \frac{t_{1}^{2}}{t_{2}^{2}} \\ \frac{CB}{AC} & \ = \pm \frac{t_{2}}{t_{1}} \Rightarrow \frac{CB}{AC} + 1 = \pm \frac{t_{2}}{t_{1}} + 1 \\ \Rightarrow \ \frac{AB}{AC} & \ = \frac{t_{2} + t_{1}}{t_{1}} = 3 \\ \text{~}\text{or}\text{~}\ \frac{AB}{AC} & \ = \frac{t_{1} - t_{2}}{t_{1}} = 3 \end{matrix} \end{matrix}$$

$$\begin{matrix} \Rightarrow & t_{2} + t_{1} = 3t_{1}\text{~}\text{or}\text{~}t_{1} - t_{2} = 3t_{1} \\ \Rightarrow & 2t_{1} - t_{2} = 0\text{~}\text{or}\text{~}2t_{1} + t_{2} = 0 \\ & \left\lbrack \because\text{~}\text{if}\text{~}\frac{CB}{AC} = - \frac{t_{2}}{t_{1}} \Rightarrow t_{1}\text{~}\text{and}\text{~}t_{2} \right.\ \text{~}\text{are of opp.}\text{ }\text{sign.}\text{~}\rbrack\text{~} \\ & \text{~} \\ & 6t_{1}^{2} - 2t_{2}^{2} - t_{1}t_{2} = 0 = 6t_{1}^{2} + 3t_{1}t_{2} - 4t_{1}t_{2} - 2t_{2}^{2} \\ & \\ \Rightarrow & \left( 3t_{1} - 2t_{2} \right)\left( 2t_{1} + t_{2} \right) = 0 \end{matrix}$$