Gaokao Math Practice Question
已知椭圆 $T: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \left(a>b>0\right)$ 的离心率为 $\frac{\sqrt{3}}{2}$, 过右焦点 $F$ 且 斜率为 $k(k>0)$ 的直线与 $T$ 相交于 $A, B$ 两点, 若 $\overline{\mathrm{AF}}=3 \overline{\mathrm{FB}}$, 则 $k=$ ($\qquad$)
- A.1
- B.$\sqrt{2}$
- C.$\sqrt{3}$
- D.2
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