所属成套资源:2024年高考数学第一轮复习专题训练(附单独答案解析)
2024年数学高考大一轮复习第十二章 §12.2 参数方程(附答单独案解析)
展开这是一份2024年数学高考大一轮复习第十二章 §12.2 参数方程(附答单独案解析),共6页。试卷主要包含了了解参数方程,了解参数的意义等内容,欢迎下载使用。
§12.2 参数方程
考试要求 1.了解参数方程,了解参数的意义.2.能选择适当的参数写出直线、圆和椭圆的参数方程.
知识梳理
1.参数方程和普通方程的互化
(1)一般地,在平面直角坐标系中,如果曲线上任意一点的坐标x,y都是某个变数t的函数并且对于t的每一个允许值,由方程组所确定的点M(x,y)都在这条曲线上,那么此方程就叫做这条曲线的参数方程.
(2)曲线的参数方程和普通方程是曲线方程的不同形式.一般地,可以通过________________而从参数方程得到普通方程.
2.常见曲线的参数方程和普通方程
点的轨迹 | 普通方程 | 参数方程 |
直线 | y-y0=tan α·(x-x0) |
|
圆 |
| (θ为参数) |
椭圆 | +=1(a>b>0) |
|
双曲线 | -=1(a>0,b>0) | (φ为参数) |
抛物线 | y2=2px(p>0) | (t为参数) |
思考辨析
判断下列结论是否正确(请在括号中打“√”或“×”)
(1)参数方程中的x,y都是参数t的函数.( )
(2)方程(θ为参数)表示以点(0,1)为圆心,以2为半径的圆.( )
(3)已知椭圆的参数方程(t为参数),点M在椭圆上,对应参数t=,点O为原点,则直线OM的斜率为.( )
(4)参数方程(θ为参数且θ∈)表示的曲线为椭圆.( )
教材改编题
1.参数方程 (t为参数) 的图象是( )
A.离散的点 B.抛物线
C.圆 D.直线
2.参数方程 (θ为参数)化为普通方程为( )
A.x2+=1 B.x2+=1
C.y2+=1 D.y2+=1
3.已知直线l的参数方程是(t为参数),若l与圆x2+y2-4x+3=0交于A,B两点,且|AB|=,则直线l的斜率为________.
题型一 参数方程与普通方程的互化
例1 已知曲线C1,C2的参数方程为C1: (θ为参数),C2:(t为参数).
(1)将C1,C2的参数方程化为普通方程;
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(2)若点P是曲线C1上的动点,求点P到C2的距离的最小值.
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思维升华 消去方程中的参数一般有三种方法
(1)利用解方程的技巧求出参数的表达式,然后代入消去参数.
(2)利用三角恒等式消去参数.
(3)根据参数方程本身的结构特征,灵活地选用一些方法从整体上消去参数.
跟踪训练1 (2022·全国甲卷)在直角坐标系xOy中,曲线C1的参数方程为(t为参数),曲线C2的参数方程为(s为参数).
(1)写出C1的普通方程;
(2)以坐标原点为极点,x轴正半轴为极轴建立极坐标系,曲线C3的极坐标方程为2cos θ-sin θ=0,求C3与C1交点的直角坐标,及C3与C2交点的直角坐标.
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题型二 参数方程的应用
例2 在平面直角坐标系xOy中,曲线C的参数方程为 (λ为参数).
(1)求曲线C的普通方程;
(2)已知点M(2,0),直线l的参数方程为(t为参数),且直线l与曲线C交于A,B两点,求+的值.
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思维升华 (1)解决直线与曲线的参数方程的应用问题时,一般是先化为普通方程,再根据直线与曲线的位置关系来解决.
(2)对于形如(t为参数)的方程,当a2+b2≠1时,应先化为标准形式后才能利用t的几何意义解题.
跟踪训练2 (2023·榆林模拟)在直角坐标系xOy中,曲线C的参数方程为C:(t为参数),以直角坐标的原点O为极点,x轴的正半轴为极轴建立极坐标系.
(1)求曲线C的极坐标方程;
(2)若A,B是曲线C上的两点,且·=0,求||的最小值.
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题型三 极坐标方程和参数方程的综合应用
例3 (2022·全国乙卷)在直角坐标系xOy中,曲线C的参数方程为(t为参数).以坐标原点为极点,x轴正半轴为极轴建立极坐标系,已知直线l的极坐标方程为ρsin+m=0.
(1)写出l的直角坐标方程;
(2)若l与C有公共点,求m的取值范围.
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思维升华 解决参数方程和极坐标的综合问题的方法
涉及参数方程和极坐标方程的综合题,求解的一般方法是分别化为普通方程和直角坐标方程后求解.当然,还要结合题目本身特点,确定选择何种方程.
跟踪训练3 在平面直角坐标系xOy中,曲线C2的参数方程是 (α为参数),在极坐标系(与直角坐标系xOy取相同的长度单位,且以原点为极点,以x轴正半轴为极轴)中,曲线C1的极坐标方程是ρcos θ-3=0,点P是曲线C2上的动点.
(1)求点P到曲线C1的距离的最大值;
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(2)若曲线C3:θ=交曲线C2于A,B两点,求△ABC2的面积.
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