Trong mặt phẳng tọa độ Oxy cho parabol (P): y = x^2/4

$$\begin{gathered} \mathrm{trong} \mathrm{mp} \mathrm{tọa} \mathrm{độ} \mathrm{Oxy} \mathrm{cho} \mathrm{Parabol}\left(\mathrm{P}\right):\mathrm{y}=\frac{{\mathrm{x}}^2}{4}.\mathrm{Hai} \mathrm{điểm} \mathrm{A},\mathrm{B} \mathrm{thuộc} \mathrm{đồ} \mathrm{thị} \left(\mathrm{P}\right) \mathrm{lần} \mathrm{lượt} \mathrm{có} \mathrm{hoành} \mathrm{độ} \mathrm{là} 2 \mathrm{và} -4 \\ \mathrm{Hãy} \mathrm{viết} \mathrm{pt} \mathrm{đường} \mathrm{thẳng} \left(\mathrm{d}\right) \mathrm{đi} \mathrm{qua} 2 \mathrm{điểm} \mathrm{A},\mathrm{B} \end{gathered}$$