高考数学精品复习资料
2019.5
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考点44 曲线与方程、圆锥曲线的综合应用
一、选择题
1.（20xx·四川高考理科·Ｔ6）抛物线![](https://wkrtcs.bdimg.com/rtcs/image?w=53&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_16","w":"53","h":"24","dataType":"wmf","c":"word/media/image1.png","imgOriW":106,"imgOriH":48})
的焦点到双曲线的渐近线的距离是（ ）
（A）![](https://wkrtcs.bdimg.com/rtcs/image?w=16&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_16","w":"16","h":"41","dataType":"wmf","c":"word/media/image3.png","imgOriW":32,"imgOriH":82})
（B） （C） （D）
【解题指南】本题考查的是抛物线与双曲线的基本几何性质，在求解时首先求得抛物线的焦点坐标，然后求得双曲线的渐近线方程，利用点到直线的距离公式进行求解即可.
【解析】选B，由抛物线的焦点，双曲线的一条渐近线方程为，根据点到直线的距离公式可得，故选B.
2.（20xx·山东高考文科·Ｔ11）与（20xx·山东高考理科·Ｔ11）相同
抛物线C1：y=![](https://wkrtcs.bdimg.com/rtcs/image?w=27&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_16","w":"27","h":"44","dataType":"wmf","c":"word/media/image10.png","imgOriW":54,"imgOriH":88})
x2(p＞0)的焦点与双曲线C2：的右焦点的连线交C1于第一象限的点M.若C1在点M处的切线平行于C2的一条渐近线，则p=（ ）
A.![](https://wkrtcs.bdimg.com/rtcs/image?w=27&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_24","w":"27","h":"45","dataType":"wmf","c":"word/media/image12.png","imgOriW":54,"imgOriH":90})
B. C. D.
【解题指南】 本题考查了圆锥曲线的位置关系,可先将抛物线化成标准方程,然后再利用过交点的切线平行于C2的一条渐近线,求得切线斜率,进而求得p的值.
【解析】选D. 经过第一象限的双曲线的渐近线为![](https://wkrtcs.bdimg.com/rtcs/image?w=63&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_16","w":"63","h":"45","dataType":"wmf","c":"word/media/image16.png","imgOriW":126,"imgOriH":90})
.抛物线的焦点为,双曲线的右焦点为.，所以在处的切线斜率为，即，所以，即三点，，共线，所以，即.
二、填空题
3. （20xx·江西高考理科·Ｔ14）抛物线x2=2py（p＞0）的焦点为F，其准线与双曲线![](https://wkrtcs.bdimg.com/rtcs/image?w=76&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_24","w":"76","h":"44","dataType":"wmf","c":"word/media/image27.png","imgOriW":152,"imgOriH":88})
相交于A，B两点，若△ABF为等边三角形，则p=___________.
【解题指南】A、B、F三点坐标都能与p建立起联系，分析可知△ABF的高为P，可构造p的方程解决.
【解析】由题意知△ABF的高为P，将![](https://wkrtcs.bdimg.com/rtcs/image?w=52&md5sum=6902867baf6a8315f0eeadc7577c3a46&sign=95c8c6f348&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=169&ipr={"t":"img","r":"r_24","w":"52","h":"41","dataType":"wmf","c":"word/media/image28.png","imgOriW":104,"imgOriH":82})
代入双曲线方程得A，B两点的横坐标为，因为△ABF为等边三角形，所以，从而解得，即.
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