Tìm {M} biểu diễn số phức z thỏa mãn |z+1| + |z-1| = 6. {M}

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Gọi} \mathrm{M}(\mathrm{x};\mathrm{y}) \mathrm{là} \mathrm{điểm} \mathrm{biểu} \mathrm{diễn} \mathrm{số} \mathrm{phức} \mathrm{z}=\mathrm{x}+\mathrm{yi} \\ \mathrm{Ta} \mathrm{có}: \left|\mathrm{z}+1\right|+\left|\mathrm{z}-1\right|=6 \\ \iff \left|\mathrm{x}+1+\mathrm{yi}\right|+\left|\mathrm{x}-1+\mathrm{yi}\right|=6 \\ \iff \sqrt{{\left(\mathrm{x}+1\right)}^2+{\mathrm{y}}^2}+\sqrt{{\left(\mathrm{x}-1\right)}^2+{\mathrm{y}}^2}=6 \\ \mathrm{Gọi} {\mathrm{F}}_1\left(-1;0\right) \mathrm{và} {\mathrm{F}}_2\left(1;0\right) \Rightarrow {\mathrm{MF}}_1+{\mathrm{MF}}_2=6 \\ \Rightarrow \left\{\mathrm{M}\right\} \mathrm{năm} \mathrm{trên} \mathrm{elip} \left(\mathrm{E}\right) \mathrm{có} \mathrm{tiêu} \mathrm{cự} {\mathrm{F}}_1{\mathrm{F}}_2=2=2\mathrm{c}\Rightarrow \mathrm{c}=1, \mathrm{độ} \mathrm{dài} \mathrm{trục} \mathrm{lớn} 2\mathrm{a}=6\Rightarrow \mathrm{a}=3 \\ \Rightarrow \mathrm{Độ} \mathrm{dài} \mathrm{trục} \mathrm{bé}: 2\mathrm{b}=2\sqrt{{\mathrm{a}}^2-{\mathrm{c}}^2}=2\sqrt{3^2-1^2}=4\sqrt{2}\Rightarrow \mathrm{b}=2\sqrt{2} \\ \left(\mathrm{E}\right) \mathrm{có} \mathrm{phương} \mathrm{trình} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow \left\{\mathrm{M}\right\} \mathrm{là} \mathrm{elip}: \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{8}=1 \end{gathered}$$