854

已知三条直线 $l_{i} : y = k x + m_{i} (i = 1, 2, 3)$ 分别与抛物线 $Γ : y^{2} = 8 x$ 交于点 $A_{i}, B_{i}$ , $T (t, 0)$ 为 $x$ 轴上一定点，且 $m_{1} < m_{2} < m_{3} < - t$ . 记点 $T$ 到直线 $l_{i}$ 的距离为 $d_{i}$ , $Δ T A_{i} B_{i}$ 的面积为 $S_{i}$ .

若直线 $l_{3}$ 的倾斜角为 $45^{\circ}$ , 且过抛物线 $Γ$ 的焦点为 $F$ , 求直线 $l_{3}$ 的方程;
若 $\vec{O A_{1}} \cdot \vec{O A_{2}} = 0$ ，且 $k m_{1} \neq 0$ ，证明: 直线 $l_{1}$ 过定点;
当 $k = 1$ 时，是否存在点 $T$ ，使得 $S_{1}, S_{2}, S_{3}$ 成等比数列， $d_{1}, d_{2}, d_{3}$ 也成等比数列? 若存在，请求出点 $T$ 的坐标；若不存在，请说明理由.

解答

解 $F (2, 0)$ , 故 $l_{3} : y = x - 2$ .
证明联立 $l_{1}$ 与 $Γ$ 得 $\begin{array}{r} k^{2} x^{2} + 2 (k m_{i} - 4) x + m_{i}^{2} = 0. \end{array}$
由韦达定理，解 $x_{1}, x_{2}$ 满足 $\begin{array}{r} x_{1} + x_{2} = - \frac{2 (k m_{i} - 4)}{k^{2}}, x_{1} x_{2} = \frac{m_{i}^{2}}{k^{2}} . \end{array}$
记 $A_{1} (x_{A}, y_{A}), B_{1} (x_{B}, y_{B})$ .由条件，对 $i = 1$ : $\begin{aligned} 0 = & x_{A} x_{B} + y_{A} y_{B} = (k^{2} + 1) x_{A} x_{B} + k m_{1} (x_{A} + x_{B}) + m_{1}^{2} \\ = & (k^{2} + 1) \frac{m_{1}^{2}}{k^{2}} + k m_{1} \frac{8 - 2 k m_{1}}{k^{2}} + m_{1}^{2} = \frac{m_{1} (m_{1} + 8 k)}{k^{2}}, \end{aligned}$
从而 $m_{1} + 8 k = 0$ , $l_{1} : y = k x - 8 k$ ,故 $l_{1}$ 恒过 $(8, 0)$ .
解当 $k = 1$ ，由2得 $\begin{array}{r} | x_{A_{i}} - x_{B_{i}} | = \sqrt{(x_{A_{i}}^{2} + x_{B_{i}}^{2})^{2} - 4 x_{A_{i}} x_{B_{i}}} = 4 \sqrt{4 - 2 m_{i}} . \end{array}$
由点到直线距离公式得 $d_{i} = \frac{| t + m_{i} |}{\sqrt{2}}$ .则 $\begin{array}{r} d_{2}^{2} = d_{1} d_{3} \Rightarrow (t + m_{2})^{2} = (t + m_{1}) (t + m_{2}) . \end{array}$
且 $\begin{aligned} 1 = \frac{S_{2}^{2}}{S_{1} S_{3}} = \frac{A_{2} B_{2}^{2}}{A_{1} B_{1} \cdot A_{3} B_{3}} \cdot & \frac{d_{2}^{2}}{d_{1} d_{3}} = \frac{| x_{A_{2}} - x_{B_{2}} |^{2}}{| x_{A_{1}} - x_{B_{1}} | \cdot | x_{A_{3}} - x_{B_{3}} |} = \frac{4 - 2 m_{2}}{\sqrt{(4 - 2 m_{1}) (4 - 2 m_{3})}} \\ \Rightarrow & (4 - 2 m_{2})^{2} = (4 - 2 m_{1}) (4 - 2 m_{3}) . \end{aligned}$
这两式相减得 $\begin{array}{r} (t + 2) (2 m_{2} - m_{1} - m_{3}) = 0. \end{array}$
当 $2 m_{2} = m_{1} + m_{3}$ ,回代得 $m_{2}^{2} = m_{1} m_{3} = \frac{1}{4} (m_{1} + m_{3})^{2}$ ,得 $m_{1} = m_{3}$ ，不合题意；故只有 $t = - 2$ ,此时可取 $(m_{1}, m_{2}, m_{3}) = (- 1, 2 - \sqrt{3}, 1)$ 即可.