圆锥共线焦半径,其调和为半通径

圆锥共线焦半径,其调和为半通径

【引理】如图,已知\(AB//CD//EF\),\(AB = x\),\(CD = z\),\(EF = y\),\(AC = m\),\(CE = n\),则\(z = x \cdot \dfrac{n}{{m + n}} + y \cdot \dfrac{m}{{m + n}}\).

圆锥共线焦半径,其调和为半通径

【证明】设\(AE\)与\(BF\)相交于点\(P\),则由平行线分线段成比例定理得

\(\left\{ \begin{array}{l}
\dfrac{{PA + m}}{{PA}} = \dfrac{z}{x}\\
\dfrac{{PA + m + n}}{{PA}} = \dfrac{y}{x}
\end{array} \right.\)

即\(\left\{ \begin{array}{l}
\dfrac{m}{{PA}} = \dfrac{z}{x} – 1 = \dfrac{{z – x}}{x}\\
\dfrac{{m + n}}{{PA}} = \dfrac{y}{x} – 1 = \dfrac{{y – x}}{x}
\end{array} \right.\)

两式相除消掉\(PA\)得\(\dfrac{m}{{m + n}} = \dfrac{{z – x}}{{y – x}}\)

\(\therefore z = \dfrac{m}{{m + n}}(y – x) + x = \dfrac{{my + nx}}{{m + n}} = x \cdot \dfrac{n}{{m + n}} + y \cdot \dfrac{m}{{m + n}}\)

【圆锥曲线第二定义】

设\(F\)为圆锥曲线的一个焦点,直线\(l\)为对应的准线,\(e\)为圆锥曲线的离心率,\(P\)为圆锥曲线上任意一点,\(P\)到准线\(l\)的距离为\(d\),则\(\dfrac{{|PF|}}{d} = e\),也有\(d = \dfrac{{|PF|}}{e}\).

【圆锥曲线的性质】

以椭圆为例,设椭圆\(C:\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\)(\(a>b>0\)),\({F_1}\)为椭圆\(C\)的焦点,过\({F_1}\)的直线与椭圆\(C\)交于点\(P\)、\(Q\),设焦点到准线的距离(简称焦准距)为\(p = \dfrac{{{b^2}}}{c}\),则

\(\dfrac{1}{{|P{F_1}|}} + \dfrac{1}{{|Q{F_1}|}} = \dfrac{2}{{ep}}\)

即\(ep = \dfrac{c}{a} \cdot \dfrac{{{b^2}}}{c} = \dfrac{{{b^2}}}{a}\)(垂直于长轴的焦半径,通径的一半,简称为半通径)为\(|P{F_1}|\)与\(|P{F_2}|\)的调和平均数.

(注:数\(x\)与\(y\)的调和平均数为\(\dfrac{2}{{\dfrac{1}{x} + \dfrac{1}{y}}}\))

圆锥共线焦半径,其调和为半通径

【证明】如图,过\(P\)作\(PA \bot l\)于\(A\),设\(|P{F_1}| = n\),则\(|PA| = \dfrac{n}{e}\),
过\(Q\)作\(QB \bot l\)于\(B\),设\(|Q{F_1}| = m\),设\(|Q{F_1}| = m\),则\(|QB| = \dfrac{m}{e}\).

\({F_1}\)到\(l\)的距离\(|M{F_1}| = | – c – ( – \dfrac{{{a^2}}}{c})| = \dfrac{{{a^2} – {c^2}}}{c} = \dfrac{{{b^2}}}{c}\)

由于\(BQ//M{F_1}//PA\),由上述【引理】得

\(|M{F_1}| = |BQ| \cdot \dfrac{n}{{m + n}} + |PA| \cdot \dfrac{m}{{m + n}}\)

即\(\dfrac{{{b^2}}}{c} = \dfrac{m}{e} \cdot \dfrac{n}{{m + n}} + \dfrac{n}{e} \cdot \dfrac{m}{{m + n}} = \dfrac{{2mn}}{{e(m + n)}}\)

整理得\(\dfrac{{m + n}}{{mn}} = \dfrac{2}{{e \cdot \dfrac{{{b^2}}}{c}}}\),

即\(\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{2}{{ep}} = \dfrac{2}{{\dfrac{{{b^2}}}{a}}}\),得证

【例题】

设椭圆\(C\)的左右焦点为\({F_1}\)、\({F_2}\),焦距为\(2c\),过点\({F_1}\)的直线与椭圆\(C\)交于点\(P\)、\(Q\),若\(|P{F_2}| = 2c\),且\(|P{F_1}| = \dfrac{4}{3}|Q{F_1}|\),则椭圆\(C\)的离心率为(       )

A.\(\dfrac{1}{2}\)           B.\(\dfrac{3}{4}\)           C.\(\dfrac{5}{7}\)           D.\(\dfrac{2}{3}\)

【解析】

\(\left\{ \begin{array}{l}
\dfrac{1}{{|P{F_1}|}} + \dfrac{1}{{|Q{F_1}|}} = \dfrac{2}{{ep}}{\rm{ }} \cdots \cdots ①
\\
|P{F_1}| + |P{F_2}| = |P{F_1}| + 2c = 2a{\rm{ }} \cdots \cdots ②
\\
|P{F_1}| = \dfrac{4}{3}|Q{F_1}|{\rm{ }} \cdots \cdots ③
\end{array} \right.\)

由②、③得\(|P{F_1}| = 2a – 2c,|Q{F_1}| = \dfrac{3}{4}(2a – 2c)\),代入①得

\(\dfrac{1}{{2a – 2c}} + \dfrac{4}{{3(2a – 2c)}} = \dfrac{2}{{\dfrac{c}{a} \cdot \dfrac{{{b^2}}}{c}}}\)

即\(\dfrac{7}{{3(2a – 2c)}} = \dfrac{{2a}}{{{b^2}}} = \dfrac{{2a}}{{{a^2} – {c^2}}}\)

整理得\(7({a^2} – {c^2}) = 12({a^2} – ac)\)

等式两边同除以\({a^2}\)得\(7(1 – {e^2}) = 12(1 – e)\)

整理得\(7{e^2} – 12e + 5 = 0\),解得\(e = \dfrac{5}{7}\),故选C

原创文章,作者:leopold,如若转载,请注明出处:https://www.math211.com/2021/02/10/88/

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