解法一:設(shè)橢圓與雙曲線的交點(diǎn)為
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,由橢圓、雙曲線定義,及已知條件得:
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或
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即
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化簡得
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或
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即:
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化簡得:
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∴所求軌跡方程為
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軌跡是兩個(gè)圓除去與
y軸的交點(diǎn)。
解法二:由題意設(shè)雙曲線的實(shí)半軸長為
a
則橢圓的半長就是
a又∵
c =" 4 "
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為橢圓半短軸
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為雙曲線的虛軸
則橢圓方程為
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……(1)
雙曲線方程為
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……(2)
由(1)×4-(2)得
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即
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……(3)
(3)代入(2)得:
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代回(2)中消去
a得
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若
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即
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即
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則所求的軌跡是兩個(gè)圓除去它們與
y軸的交點(diǎn),方程是:
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通過橢圓和雙曲線定義,建立動(dòng)點(diǎn)滿足的幾何條件,再坐標(biāo)化而得到軌跡方程。
或由焦點(diǎn)已知曲線中收為原點(diǎn),坐標(biāo)軸為對稱軸,再需一個(gè)條件用待定系法也可求軌跡方程。解法一是將“a”當(dāng)作參數(shù)引進(jìn)后來后建立方程,不如解法一直接使用定義尋找到動(dòng)點(diǎn)滿足的幾何關(guān)系簡單。
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
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2
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