离心率:直线参数方程的应用2例

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直线的参数方程可以很方便地表示直线上的点到定点的距离,下面以两例说明其应用。


1. 已知F_1F_2是双曲线\Gamma : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \, (a>0,b>0)的左右焦点,过F_1的直线l交双曲线的两条渐近线于AB两点,且|F_2A| = |F_2B||OA||AB||OB|成等比数列(公比不为1),则双曲线的离心率为(   )

A.\sqrt2     B.\sqrt3     C.2\sqrt2     D.2\sqrt3


    设直线l\begin{cases}x = -c + t\cos \theta \\ y = t\sin \theta \end{cases},与\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0联立得

    \[ \left( \dfrac{\cos ^2 \theta}{a^2} - \dfrac{\sin ^2 \theta}{b^2} \right) t^2 - \dfrac{2c}{a^2}\cos \theta \cdot t + \dfrac{c^2}{a^2} = 0 \cdots (1)\]

所以

    \[ t_1 + t_2 = \dfrac{2c \cdot \cos \theta}{\cos ^2 \theta - \dfrac{a^2 \sin^2 \theta}{b^2}} \cdots (2) \]

AB的中点M,则由|F_2A| = |F_2B|F_2M \bot F_1M,故

    \[ F_1M = \dfrac{t_1 + t_2}{2} = 2c \cdot \cos \theta \cdots (3) \]

由(2)(3)得

    \[ \sin\theta = \dfrac{b}{\sqrt2 c} ; \,\, \cos \theta = \dfrac{\sqrt{a^2+c^2}}{\sqrt2 c} \]

代回(1)式得

    \[ t^2 - 2\sqrt2 \sqrt{a^2 + c^2}t + 2c^2 = 0 \]

所以

    \[ AB = |t_2 - t_1| = 2\sqrt2 a, \,\, MF_2 = 2c\sin\theta = \sqrt2 b \]

易知S_{\triangle ABF_2} = 2S_{\triangle ABO},且有OA \cdot OB = AB^2,设\angle BOy = \alpha,有

    \[ \dfrac12 \times AB \times MF_2 = 2 \times \dfrac12 \times OA \times OB \times \sin 2\alpha = AB^2 \times \sin 2\alpha \]

代入ABMF_2

    \[ \sin 2\alpha = \dfrac{b}{4a} \]

\sin \alpha = \dfrac{a}{c}\cos \alpha = \dfrac{b}{c},所以

    \[ \dfrac{b}{4a} = 2 \cdot \dfrac{a}{c} \cdot \dfrac{b}{c}\]

解得e = 2\sqrt2


2. 如图,双曲线\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\,(a>0,b>0)的右顶点为A,左右焦点分别为F_1F_2,点P是双曲线右支上一点,PF_1交左支于点Q,交渐近线y = \dfrac{b}{a}x于点RMPQ的中点,若RF_2 \bot PF_1,且AM \bot PF_1,则双曲线的离心率是(   )


    设直线PQ\begin{cases} x = -c + t \cos \theta \\ y = t \sin \theta \end{cases},由RF_1 \bot RF_2OR = OF_1\angle ROF_2 = 2\theta。将直线PQ的参数方程与双曲线联立得:

    \[ \left( \dfrac{\cos ^2 \theta}{a^2} - \dfrac{\sin ^2 \theta}{b^2} \right) t^2 - \dfrac{2c}{a^2}\cos \theta \cdot t + \dfrac{b^2}{a^2} = 0 \]

所以

    \[ t_1 + t_2 = \dfrac{2c \cdot \cos \theta}{\cos ^2 \theta - \dfrac{a^2 \sin^2 \theta}{b^2}} \cdots (1)\]

MF_1 \bot MA

    \[ MF_1 = \dfrac{t_1+t_2}{2} = (a+c)\cdot \cos \theta \cdots (2) \]

由(1)(2)解得

    \[ \sin ^2 \theta = \dfrac{a(c-a)}{c^2} \cdots (3)\]

    \[ \sin ^2 \theta = \dfrac{1- \cos 2\theta}{2} = \dfrac{1-\dfrac{a}{c}}{2} = \dfrac{c-a}{2c} \]

与(3)式联立解得e = 2

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