宝鸡中学2017届高三总复习同步微专题39:周考-圆锥曲线定点问题

求证直线是否过定点,一般先求出直线方程,让然再看是否过定点。以一道宝鸡中学2017届周考题为例说明。

1. 已知抛物线C:y^2 = 2px\,(p>0)与椭圆C':\dfrac{x^2}{4} + \dfrac{15y^2}{16} = 1相交所得的弦长为2p

(1) 求抛物线C的标准方程;

(2) 设A,B是抛物线C上异于原点O的两个不同点,直线OAOB的倾斜角分别为\alpha\beta,当\alpha,\beta变化且\alpha + \beta为定值\theta (\tan \theta = 2)时,证明:直线AB恒过定点,并求出该定点的坐标。

分析与解    (1) 我们知道抛物线的通径长为2p,由抛物线与椭圆的对称性知,CC'的相交弦就是抛物线的通径,即CC'的交点坐标为\left( \dfrac{p}{2},p \right),代入椭圆方程解得p = 1,故抛物线方程为y^2 = 2x

(2) 设A(x_1,y_1)B(x_2,y_2),则y_1^2 = 2x_1y_2^2 = 2x_2\therefore \dfrac{y_1}{x_1} = \dfrac{2}{y_1}\dfrac{y_2}{x_2} = \dfrac{2}{y_2}

\therefore \tan \theta = \tan(\alpha + \beta)

= \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \dfrac{\dfrac{y_1}{x_1} + \dfrac{y_2}{x_2}}{1 - \dfrac{y_1}{x_1}\dfrac{y_2}{x_2}}

= \dfrac{\dfrac{2}{y_1} + \dfrac{2}{y_2}}{1 - \dfrac{2}{y_1} \dfrac{2}{y_2}}

= \dfrac{2(y_1 + y_2)}{y_1y_2 - 4} = 2

\therefore y_1+y_2 = y_1y_2 - 4 \cdots \cdots (1)

直线AB的方程为\dfrac{y - y_1}{x - x_1} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{y_2 - y_1}{\dfrac{y_2^2}{2} - \dfrac{y_1^2}{2}} =\dfrac{2}{y_1 + y_2}

(y_1 + y_2)(y - y_1) = 2(x - x_1),展开化简得2x - (y_1+y_2)y + y_1y_2 = 0

结合(1)式知,AB恒过点(-2,1)

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