解析几何条件翻译案例

一、解析几何该如何思考？思考什么？在解析几何中必须要思考和解决的基本且关键的问题：

问题1：显性条件和目标信息如何翻译？

如\(OA \bot OB\)翻译成：\({k_{OA}} \cdot {k_{OB}} = – 1\)（斜率是否一定存在？）或​\( \overrightarrow{OA} \bullet\overrightarrow{OB}=0 \)​

问题2：是否存在隐形条件及关系？

问题3：目标信息“直线过定点”如何表达？

问题4：需引入哪些参数？

引参情形一：设点

引参情形二：设点、设线结合

问题5：如何假设设直线？

若直线\(l\)与抛物线\({y^2} = 2px\)有两个焦点，则隐含着直线一定不能平行于\(x\)轴，但可以垂直与\(x\)轴，于是反设直线\(l\)：\(x = ty + m\)是不错的选择．

反设直线的特征为：

① 核心条件与目标信息形式为\(y\)为主；

② 直线过\(x\)轴上的定点．

【注】直线的不同形式对后续运算影响很大．

问题6：联立后是消\(x\)还是消\(y\)？

保留谁，最终都是服务于目标信息的，需要根据核心信息与目标信息加以分析．

问题7：核心信息坐标化后如何韦达化？

问题8：坐标的代换用直线还是曲线？

椭圆一般用直线代换，抛物线有时用曲线代换更简单，如\({x_1}{x_2} = \dfrac{{y_1^2}}{{2p}} \cdot \dfrac{{y_2^2}}{{2p}} = \dfrac{{{{({y_1}{y_2})}^2}}}{{4{p^2}}}\)

二、核心条件的坐标化翻译，是解题的关键之一．

部分转化示例：

（一）\(\dfrac{{|A{F_1}|}}{{|B{F_1}|}} = \dfrac{{\sqrt {1 + {k^2}} |{x_1} – {x_F}|}}{{\sqrt {1 + {k^2}} |{x_2} – {x_F}|}} = \dfrac{{|{x_1} + c|}}{{|{x_2} + c|}}\)；\[\dfrac{{|A{F_1}|}}{{|B{F_1}|}} = \dfrac{{\sqrt {1 + {t^2}} |{y_1} – {y_F}|}}{{\sqrt {1 + {t^2}} |{y_2} – {y_F}|}} = \dfrac{{|{y_1}|}}{{|{y_2}|}}\]
\(|A{F_1}||B{F_1}| = – \overrightarrow {A{F_1}} \bullet \overrightarrow {B{F_1}} \)；\(|A{F_1}||B{F_1}| = (a + e{x_1})(a + e{x_2})\)

【案例】椭圆\(C:\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{2} = 1\)，​\( F_1 \)​为椭圆左焦点，是否存在实数\(t\)，使得
\(|A{F_1}| + |B{F_1}| = t|A{F_1}||B{F_1}|\)恒成立？若存在，求出\(t\)的值；若不存在，说明理由．
【分析】准备阶段：当直线\(AB\)斜率不为零时，设直线\(AB:x = my – \sqrt 2 \)，代入\({x^2} + 2{y^2} = 4\)得\(({m^2} + 2){y^2} – 2\sqrt 2 my – 2 = 0\)，\(\Delta = 16({m^2} + 1)>0\)，设\(A({x_1},{y_1})\)、\(B({x_2},{y_2})\)，则
\({y_1} + {y_2} = \dfrac{{2\sqrt 2 m}}{{{m^2} + 2}},{y_1}{y_2} = – \dfrac{2}{{{m^2} + 2}}\)
法一：利用距离公式转化
\(t = \dfrac{{|A{F_1}| + |B{F_1}|}}{{|A{F_1}||B{F_1}|}} = \dfrac{{\sqrt {1 + {m^2}} |{y_1} – {y_2}|}}{{\sqrt {1 + {m^2}} |{y_1}| \cdot \sqrt {1 + {m^2}} |{y_2}|}} = \dfrac{{\dfrac{{\sqrt {16({m^2} + 1)} }}{{{m^2} + 2}}}}{{\sqrt {1 + {m^2}} \cdot |\dfrac{{ – 2}}{{{m^2} + 2}}|}} = 2\)
法二：利用焦半径公式
\(t = \dfrac{{|A{F_1}| + |B{F_1}|}}{{|A{F_1}||B{F_1}|}} = \dfrac{{a + e{x_1} + a + e{x_2}}}{{(a + e{x_1})(a + e{x_2})}} = \dfrac{{4 + \dfrac{{\sqrt 2 }}{2}({x_1} + {x_2})}}{{4 + \sqrt 2 ({x_1} + {x_2}) + \dfrac{1}{2}{x_1}{x_2}}}\)
\( = \dfrac{{4 + \dfrac{{\sqrt 2 }}{2}[m({y_1} + {y_2}) – 2\sqrt 2 ]}}{{4 + \sqrt 2 [m({y_1} + {y_2}) – 2\sqrt 2 ] + \dfrac{1}{2}(m{y_1} – \sqrt 2 )(m{y_2} – \sqrt 2 )}} = \dfrac{{\dfrac{{\sqrt 2 }}{2}m({y_1} + {y_2}) + 2}}{{\dfrac{1}{2}{m^2}{y_1}{y_2} + \dfrac{{\sqrt 2 }}{2}m({y_1} + {y_2}) + 1}}\)
\( = \dfrac{{\dfrac{{\sqrt 2 }}{2}m \cdot \dfrac{{2\sqrt 2 m}}{{{m^2} + 2}} + 2}}{{\dfrac{1}{2}{m^2} \cdot \dfrac{{ – 2}}{{{m^2} + 2}} + \dfrac{{\sqrt 2 }}{2}m \cdot \dfrac{{2\sqrt 2 m}}{{{m^2} + 2}} + 1}} = \dfrac{{2{m^2} + 2{m^2} + 4}}{{ – {m^2} + 2{m^2} + {m^2} + 2}} = 2\)

（二）非对称韦达式

【类型一】\(\dfrac{{{x_1}}}{{{x_2}}}\)比值类

方法：比值类，取倒数，再相加

【例1】已知椭圆\(C:\dfrac{{{x^2}}}{2} + {y^2} = 1\)，过点\(P(0,2)\)的直线\(l\)交\(C\)于\(A\)、\(B\)两点（点\(B\)在\(A\)、\(P\)之间），且有\(\overrightarrow {PB} = \lambda \overrightarrow {PA} \)，求\(\lambda \)的取值范围．

【分析】准备阶段：

​\( \begin{cases} y=kx+2 \\ x^2+2y^2=2 \end{cases}\Rightarrow(1+2k^2)x^2+8kx+6=0 \)​

\(\Delta = 16{k^2} – 24>0 \Rightarrow k< – \dfrac{{\sqrt 6 }}{2}\)或\(k>\dfrac{{\sqrt 6 }}{2}\)，设\(A({x_1},{y_1}),B({x_2},{y_2})\)
\({x_1} + {x_2} = – \dfrac{{8k}}{{1 + 2{k^2}}},{x_1}{x_2} = \dfrac{6}{{1 + 2{k^2}}}\)
\(\overrightarrow {PB} = \lambda \overrightarrow {PA} \Leftrightarrow ({x_2},{y_2} – 2) = \lambda ({x_1},{y_1} – 2)\)，利用\({x_2} = \lambda {x_1}\)
\(\lambda + \dfrac{1}{\lambda } = \dfrac{{{x_2}}}{{{x_1}}} + \dfrac{{{x_1}}}{{{x_2}}} = \dfrac{{{{({x_1} + {x_2})}^2} – 2{x_1}{x_2}}}{{{x_1}{x_2}}} = \dfrac{{{{( – \dfrac{{8k}}{{1 + 2{k^2}}})}^2}}}{{\dfrac{6}{{1 + 2{k^2}}}}} – 2 = \dfrac{{32}}{{\dfrac{3}{{{k^2}}} + 6}} – 2\)
令\(t = {k^2}\)，则\(g(t) = \dfrac{{32}}{{\dfrac{3}{t} + 6}} – 2\)在\((\dfrac{3}{2}, + \infty )\)上单调递增，从而\(2<\lambda + \dfrac{1}{\lambda }<\dfrac{{10}}{3}\)
注意到\(0<\lambda <1\)，解得\(\dfrac{1}{3}<\lambda <1\)
当直线\(l\)斜率不存在时，\(\lambda = \dfrac{{|PB|}}{{|PA|}} = \dfrac{1}{3}\)
综上所述， \(\lambda \)的取值范围是\([\dfrac{1}{3},1)\)

【类型二】\({y_2} + 7{y_1} = 8\)系数不等且带常数类

方法：系数不等类，按数列方法凑常数比值，再取倒数相加

【例2】已知抛物线\({y^2} = 4x\)，过焦点\(F\)的直线\(l\)与抛物线交于\(A\)、\(B\)两点，若\(\overrightarrow {AF} = 3\overrightarrow {FB} \)，则直线\(l\)的斜率为___________.

【分析】设\(A({x_1},{y_1}),B({x_2},{y_2})\)
解法一（常规解法）翻译条件\(\overrightarrow {AF} = 3\overrightarrow {FB} \)\( \Leftrightarrow (1 – {x_1}, – {y_1}) = 3({x_2} – 1,{y_2})\)
利用\({y_1} = – 3{y_2}\)，比值取倒数相加：
\( – 3 – \dfrac{1}{3} = \dfrac{{{y_1}}}{{{y_2}}} + \dfrac{{{y_2}}}{{{y_1}}} = \dfrac{{{{({y_1} + {y_2})}^2}}}{{{y_1}{y_2}}} – 2\)
接下来设直线\(l:x = my + 1\)，代入\({y^2} = 4x\)得\({y^2} – 4my – 4 = 0\)，\({y_1} + {y_2} = 4m,{y_1}{y_2} = – 4\)
代入上式得\( – \dfrac{{10}}{3} = \dfrac{{16{m^2}}}{{ – 4}} – 2\)，解得\(m = \pm \dfrac{{\sqrt 3 }}{3}\)，从而直线\(l\)的斜率为\(\dfrac{1}{m} = \pm \sqrt 3 \)
解法二（几何方法）
设直线的倾斜角为\(\theta \)，利用抛物线的定义，可构造直角梯形，然后分成一个矩形和一个直角三角形，在直角三角形中有结论\(|\cos \theta | = |\dfrac{{\lambda – 1}}{{\lambda + 1}}| = \dfrac{1}{2} \Rightarrow \theta = 60^\circ \)或\(120^\circ \)
斜率为\( \pm \sqrt 3 \)

【例3】已知抛物线\({y^2} = 4x\)与定点\(P(2,1)\)，直线\(l\)与抛物线交于\(A\)、\(B\)两点，且有\(\overrightarrow {AP} = \dfrac{1}{7}\overrightarrow {PB} \)，求直线\(l\)的斜率．

【分析】设\(A({x_1},{y_1}),B({x_2},{y_2})\)
翻译条件：\(\overrightarrow {AP} = \dfrac{1}{7}\overrightarrow {PB} \Leftrightarrow (2 – {x_1},1 – {y_1}) = \dfrac{1}{7}({x_2} – 2,{y_2} – 1)\)
得到两个关系：\(7{x_1} + {x_2} = 16\)，\(7{y_1} + {y_2} = 8\)
依然选择\(y\)来处理，如何凑出韦达式呢？首先配出比例
类比处理数列\({a_{n + 1}} = – 7{a_n} + 8\)
利用不动点法：设\(x = – 7x + 8\)，两式相减得\({a_{n + 1}} – x = – 7({a_n} – x)\)，此处\(x = 1\)
于是\(7({y_1} – 1) = – ({y_2} – 1)\)
再用取倒相加得\(( – 7) + ( – \dfrac{1}{7}) = \dfrac{{{y_2} – 1}}{{{y_1} – 1}} + \dfrac{{{y_1} – 1}}{{{y_2} – 1}} = \dfrac{{{{({y_1} + {y_2} – 2)}^2}}}{{({y_1} – 1)({y_2} – 1)}} – 2 = \dfrac{{{{({y_1} + {y_2} – 2)}^2}}}{{{y_1}{y_2} – ({y_1} + {y_2}) + 1}} – 2\)
设直线\(l:y – 1 = k(x – 2)\)，将抛物线代入直线得\(\dfrac{k}{4}{y^2} – y + 1 – 2k = 0\)
\({y_1} + {y_2} = \dfrac{4}{k},{y_1}{y_2} = \dfrac{{4(1 – 2k)}}{k}\)，代入上式得\( – \dfrac{{50}}{7} = \dfrac{{{{(\dfrac{4}{k} – 2)}^2}}}{{\dfrac{{4(1 – 2k)}}{k} – \dfrac{4}{k} + 1}} – 2 = – \dfrac{{{{(\dfrac{4}{k} – 2)}^2}}}{7} – 2\)
解得\(k = \dfrac{1}{2}\)或\(k = – 1\)

【类型三】\(\dfrac{{k{x_1}{x_2} – {x_1} + 3{x_2}}}{{k{x_1}{x_2} – 2{x_2} + {x_1}}}\)分式上下不对称类

方法：证明阻碍类，观察韦达，用两根积与两根之和互相替换

【例4】已知椭圆\(C:\dfrac{{{x^2}}}{2} + {y^2} = 1\)，\(F\)为\(C\)的右焦点，过\(F\)且不与\(x\)轴重合的直线\(l\)交\(C\)于\(A\)、\(B\)两点，过点\(A\)作直线\(x = 2\)的垂线，垂足为\(D\)，证明：直线\(BD\)过定点\(E\)．

【分析】设\(A({x_1},{y_1}),B({x_2},{y_2})\)
准备阶段：设直线\(l:x = my + 1\)，代入\({x^2} + 2{y^2} = 2\)得
\(({m^2} + 2){y^2} + 2my – 1 = 0\)，\({y_1} + {y_2} = – \dfrac{{2m}}{{{m^2} + 2}},{y_1}{y_2} = – \dfrac{1}{{{m^2} + 2}}\)
写出直线\(BD\)的方程：\(y – {y_1} = \dfrac{{{y_2} – {y_1}}}{{{x_2} – 2}}(x – 2)\)
由点\(A\)、\(B\)的对称互换性可知，若直线\(BD\)过定点，则定点必定在\(x\)轴上
令\(y = 0\)得，\(x = 2 – \dfrac{{{y_1}({x_2} – 2)}}{{{y_2} – {y_1}}} = 2 – \dfrac{{m{y_1}{y_2} – {y_1}}}{{{y_2} – {y_1}}}\)
该式中，\({y_1}{y_2}\)造成了证明的障碍，导致非其次式不对称
由韦达定理有关系式：\({y_1} + {y_2} = 2m{y_1}{y_2}\)
从而\(x = 2 – \dfrac{{m{y_1}{y_2} – {y_1}}}{{{y_2} – {y_1}}} = 2 – \dfrac{{\dfrac{1}{2}({y_1} + {y_2}) – {y_1}}}{{{y_2} – {y_1}}} = \dfrac{3}{2}\)
即直线\(BD\)恒过定点\(E(\dfrac{3}{2},0)\)

【例5】椭圆\(C:\dfrac{{{x^2}}}{6} + \dfrac{{{y^2}}}{4} = 1\)与定点\(A(0, – 2)\)，经过点\(E(0,1)\)且斜率存在的直线\(l\)交椭圆于\(Q\)、\(N\)两点，点\(B\)与点\(Q\)关于坐标原点对称，连接\(AB\)、\(AN\)，求证：存在实数\(\lambda \)，使\({k_{AN}} = \lambda {k_{AB}}\)成立．

【分析】设\(Q({x_1},{y_1}),N({x_2},{y_2})\)，则\(B( – {x_1}, – {y_1})\)
\({k_{AN}} = \dfrac{{{y_2} + 2}}{{{x_2}}},{k_{AB}} = \dfrac{{ – 2 + {y_1}}}{{{x_1}}}\)
\(\lambda = \dfrac{{{k_{AN}}}}{{{k_{AB}}}} = \dfrac{{({y_2} + 2){x_1}}}{{{x_2}({y_1} – 2)}}\)
设直线\(l:y = kx + 1\)，代入\(2{x^2} + 3{y^2} = 12\)得\((2 + 3{k^2}){x^2} + 6kx – 9 = 0\)，
\({x_1} + {x_2} = – \dfrac{{6k}}{{2 + 3{k^2}}},{x_1}{x_2} = – \dfrac{9}{{2 + 3{k^2}}}\)
\(\lambda = \dfrac{{(k{x_1} + 3){x_1}}}{{{x_2}(k{x_1} – 1)}} = \dfrac{{k{x_1}{x_2} + 3{x_1}}}{{k{x_1}{x_2} – {x_2}}}\)
此式非韦达定理对称式，无法直接代入韦达定理，
由于\(3({x_1} + {x_2}) = 2k{x_1}{x_2}\)，故\(\lambda = \dfrac{{\dfrac{3}{2}({x_1} + {x_2}) + 3{x_1}}}{{\dfrac{3}{2}({x_1} + {x_2}) – {x_2}}} = \dfrac{{\dfrac{9}{2}{x_1} + \dfrac{3}{2}{x_2}}}{{\dfrac{3}{2}{x_1} + \dfrac{1}{2}{x_2}}} = 3\)
此法的关键在于将二次的\({x_1}{x_2}\)化为一次的\({x_1} + {x_2}\)