椭圆三点定与动,所得三弦有联系

椭圆三点定与动,所得三弦有联系。

斜率之积为定值,定点对弦过定点;

斜积负一为特情,张直角弦过定点。

斜率和定非零时,定点对弦过定点;

斜率之和为零时,定点对弦斜率定。

设\(P({x_0},{y_0})\)为椭圆\(C:\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\)(\(a>b>0\))上一定点,\(A({x_1},{y_1})\)、\(B({x_2},{y_2})\)为椭圆\(C\)上两动点.

(1)若\({k_{PA}} \cdot {k_{PB}} = t(t \ne \dfrac{b^2}{a^2})\)为定值,则直线\(AB\)恒过定点;

           若\({k_{PA}} \cdot {k_{PB}} = \dfrac{b^2}{a^2}\),则直线\(AB\)的斜率为定值\(-\dfrac{y_0}{x_0}\);

(2)若\({k_{PA}} + {k_{PB}} = s(s \ne 0)\)为定值,则直线\(AB\)恒过定点;

(3)若\({k_{PA}} + {k_{PB}} = 0\),则直线\(AB\)的斜率为定值.

【解析】设直线\(AB:y = kx + m\),代入椭圆\({b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}\)得

\(({b^2} + {a^2}{k^2}){x^2} + 2km{a^2}{x^2} + {a^2}{m^2} – {a^2}{b^2} = 0\)

有韦达定理

\({x_1} + {x_2} = – \dfrac{{2km{a^2}}}{{{b^2} + {a^2}{k^2}}}\),\({x_1}{x_2} = \dfrac{{{a^2}{m^2} – {a^2}{b^2}}}{{{b^2} + {a^2}{k^2}}}\)

(1)\({k_{PA}} \cdot {k_{PB}} = \dfrac{{{y_1} – {y_0}}}{{{x_1} – {x_0}}} \cdot \dfrac{{{y_2} – {y_0}}}{{{x_2} – {x_0}}} = \dfrac{{k{x_1} + m – {y_0}}}{{{x_1} – {x_0}}} \cdot \dfrac{{k{x_2} + m – {y_0}}}{{{x_2} – {x_0}}} = t\)

整理得

\(({k^2} – t){x_1}{x_2} + (km – k{y_0} + t{x_0})({x_1} + {x_2}) + {(m – {y_0})^2} – tx_0^2 = 0\)

韦达定理代入得\(({k^2} – t)\dfrac{{{a^2}{m^2} – {a^2}{b^2}}}{{{b^2} + {a^2}{k^2}}} + (km – k{y_0} + t{x_0}) \cdot \dfrac{{ – 2km{a^2}}}{{{b^2} + {a^2}{k^2}}} + {m^2} – 2m{y_0} + y_0^2 – tx_0^2 = 0\)

去分母展开:

\({k^2}{a^2}{m^2} – t{a^2}{m^2} – {k^2}{a^2}{b^2} + t{a^2}{b^2} – 2{k^2}{m^2}{a^2} + 2{k^2}m{a^2}{y_0} – 2km{a^2}t{x_0}\)

\( + {b^2}{m^2} – 2m{b^2}{y_0} + {b^2}y_0^2 – t{b^2}x_0^2 + {a^2}{k^2}{m^2} – 2m{a^2}{k^2}{y_0} + {a^2}{k^2}y_0^2 – {a^2}{k^2}tx_0^2 = 0\)

整理得:

\( – t{a^2}{m^2} – {k^2}{a^2}{b^2} + t{a^2}{b^2} – 2km{a^2}t{x_0} + {b^2}{m^2} – 2m{b^2}{y_0} \\+ {b^2}y_0^2 – t{b^2}x_0^2 + {a^2}{k^2}y_0^2 – {a^2}{k^2}tx_0^2 = 0\)

利用主元法分解因式,把\(k\)作为主元,按\(k\)降幂整理

\(({a^2}y_0^2 – {a^2}x_0^2t – {a^2}{b^2}){k^2} – 2m{a^2}t{x_0}k + ({b^2} – t{a^2}){m^2} -\\ 2m{b^2}{y_0} + t{a^2}{b^2} + {b^2}y_0^2 – t{b^2}x_0^2 = 0\)

注意到:\({b^2}x_0^2 + {a^2}y_0^2 = {a^2}{b^2}\),故\({a^2}y_0^2 – {a^2}x_0^2t – {a^2}{b^2} = – (t{a^2} + {b^2})x_0^2\),

\(t{a^2}{b^2} + {b^2}y_0^2 – t{b^2}x_0^2 = (t{a^2} + {b^2})y_0^2\)

从而\( – (t{a^2} + {b^2})x_0^2{k^2} – 2m{a^2}t{x_0}k + ({b^2} – t{a^2}){m^2} – 2m{b^2}{y_0} + (t{a^2} + {b^2})y_0^2 = 0\)

十字相乘法分解因式:

\[\begin{array}{*{20}{c}}
1\\
{{b^2} – t{a^2}}
\end{array}\xcancel{{}}\begin{array}{*{20}{c}}
{ – {y_0}}\\
{ – (t{a^2} + {b^2}){y_0}}
\end{array}\]

故\( – (t{a^2} + {b^2})x_0^2{k^2} – 2m{a^2}t{x_0}k + (m – {y_0})[({b^2} – t{a^2})m – (t{a^2} + {b^2}){y_0}] = 0\)

再用十字相乘法分解因式:

\[\begin{array}{*{20}{c}}
{ – {x_0}}\\
{(t{a^2} + {b^2}){x_0}}
\end{array} \xcancel{{}} {{\begin{array}{*{20}{c}}
{}\\
{}
\end{array}}}\begin{array}{*{20}{c}}
{ – (m – {y_0})}\\
{ – [({b^2} – t{a^2})m – (t{a^2} + {b^2}){y_0}]}
\end{array}\]

从而有\(( – {x_0}k – m + {y_0})[(t{a^2} + {b^2}){x_0}k – ({b^2} – t{a^2})m + (t{a^2} + {b^2}){y_0}] = 0\)

注意到直线\(y = kx + m\)不过点\(P({x_0},{y_0})\),故\({y_0} \ne k{x_0} + m\)

从而\((t{a^2} + {b^2}){x_0}k – ({b^2} – t{a^2})m + (t{a^2} + {b^2}){y_0} = 0\),

当\({b^2} – t{a^2}=0\),即\(t=\dfrac{b^2}{a^2}\)时,\(k=-\dfrac{y_0}{x_0}\)

当\({b^2} – t{a^2} \ne 0\),即\(t \ne \dfrac{b^2}{a^2}\)时,

解得\[m = \dfrac{{(t{a^2} + {b^2}){x_0}}}{{{b^2} – t{a^2}}}k + \dfrac{{(t{a^2} + {b^2}){y_0}}}{{{b^2} – t{a^2}}}\]

所以直线\(AB\)的方程可化为:
\[y = k[x + \dfrac{{(t{a^2} + {b^2}){x_0}}}{{{b^2} – t{a^2}}}] + \dfrac{{(t{a^2} + {b^2}){y_0}}}{{{b^2} – t{a^2}}}\]

从而直线\(AB\)恒过定点\[\left( { – \dfrac{{(t{a^2} + {b^2}){x_0}}}{{{b^2} – t{a^2}}},\dfrac{{(t{a^2} + {b^2}){y_0}}}{{{b^2} – t{a^2}}}} \right)\]

特别地,当\({k_{PA}} \cdot {k_{PB}} = t = – 1\)即\(PA \bot PB\)时,直线\(AB\)恒过定点\[\left( {\dfrac{{{c^2}{x_0}}}{{{b^2} + {a^2}}},\dfrac{{ – {c^2}{y_0}}}{{{b^2} + {a^2}}}} \right)\]
(2)\[{k_{PA}} + {k_{PB}} = \dfrac{{{y_1} – {y_0}}}{{{x_1} – {x_0}}} + \dfrac{{{y_2} – {y_0}}}{{{x_2} – {x_0}}} = \dfrac{{k{x_1} + m – {y_0}}}{{{x_1} – {x_0}}} + \dfrac{{k{x_2} + m – {y_0}}}{{{x_2} – {x_0}}}\]

其中

\(\dfrac{{k{x_1} – k{x_0} + k{x_0} + m – {y_0}}}{{{x_1} – {x_0}}} + \dfrac{{k{x_2} – k{x_0} + k{x_0} + m – {y_0}}}{{{x_2} – {x_0}}}\)

\(= 2k + (k{x_0} + m – {y_0})\dfrac{{{x_1} + {x_2} – 2{x_0}}}{{{x_1}{x_2} – {x_0}({x_1} + {x_2}) + x_0^2}} = s,\)

\( \dfrac{{{x_1} + {x_2} – 2{x_0}}}{{{x_1}{x_2} – {x_0}({x_1} + {x_2}) + x_0^2}}\)

​\( = \dfrac{{ – \dfrac{{2km{a^2}}}{{{b^2} + {a^2}{k^2}}} – 2{x_0}}}{{\dfrac{{{a^2}{m^2} – {a^2}{b^2}}}{{{b^2} + {a^2}{k^2}}} + \dfrac{{2km{a^2}{x_0}}}{{{b^2} + {a^2}{k^2}}} + x_0^2}}\)

\( = \dfrac{{ – 2km{a^2} – 2{x_0}({b^2} + {a^2}{k^2})}}{{{a^2}{m^2} – {a^2}{b^2} + 2km{a^2}{x_0} + {b^2}x_0^2 + {a^2}x_0^2{k^2}}} \)

由于\({b^2}x_0^2 + {a^2}y_0^2 = {a^2}{b^2}\),故

\[{a^2}{m^2} – {a^2}{b^2} + 2km{a^2}{x_0} + {b^2}x_0^2 + {a^2}x_0^2{k^2} = {a^2}{m^2} – {a^2}y_0^2 + 2km{a^2}{x_0} + {a^2}x_0^2{k^2}\]

\[ = {a^2}(x_0^2{k^2} + 2km{x_0} + {m^2} – y_0^2) = {a^2}(k{x_0} + m + {y_0})(k{x_0} + m – {y_0})\]

\[\begin{array}{*{20}{c}}
{k{x_0}}\\
{k{x_0}}
\end{array} \xcancel{{}} \begin{array}{*{20}{c}}
{m + {y_0}}\\
{m – {y_0}}
\end{array}\]

所以\(2k + \dfrac{{ – 2km{a^2} – 2{x_0}({b^2} + {a^2}{k^2})}}{{{a^2}(k{x_0} + m + {y_0})}} = \dfrac{{2{a^2}k{y_0} – 2{x_0}{b^2}}}{{{a^2}{x_0}k + {a^2}m + {a^2}{y_0}}} = s\)(*)

整理得\[m = (\dfrac{{2{y_0}}}{s} – {x_0})k – \dfrac{{2{x_0}{b^2}}}{{s{a^2}}} – {y_0}\](**)

所以直线\(AB\)方程为\[y = k(x + \dfrac{{2{y_0}}}{s} – {x_0}) – \dfrac{{2{x_0}{b^2}}}{{s{a^2}}} – {y_0}\]恒过定点\(({x_0} – \dfrac{{2{y_0}}}{s}, – \dfrac{{2{x_0}{b^2}}}{{s{a^2}}} – {y_0})\)

(3)若\({k_{PA}} + {k_{PB}} = 0\),则相当于(2)中\(s\)取\(0\),故由(*)得\(k = \dfrac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\)为定值,此时(**)式没有意义,因此没有直线过定点这一结论。

原创文章,作者:leopold,如若转载,请注明出处:https://www.math211.com/2021/02/10/78/

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