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-1-2.2直线和圆的参数方程ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航1.掌握直线的参数方程及其参数的几何意义.2.能用直线的参数方程解决简单问题.3.掌握圆的参数方程及其参数的意义.4.能用圆的参数方程解决一些简单问题.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航121.直线的参数方程过定点M0(x0,y0),斜率为k的直线的点斜式方程为y-y0=k(x-x0).其中参数t的几何意义是|t|表示直线上任一点M到定点M0的距离.其中k=tanα,α为直线的倾斜角,代入上式,得y-y0=sin𝛼cos𝛼(𝑥−𝑥0),𝛼≠π2,即𝑥-𝑥0cos𝛼=𝑦-𝑦0sin𝛼.记上式的比值为𝑡,整理后得𝑥=𝑥0+𝑡cos𝛼,𝑦=𝑦0+𝑡sin𝛼,这是直线的参数方程.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12【做一做1-1】直线𝑥=-2+𝑡cos30°,𝑦=3-𝑡sin60°(𝑡为参数)的倾斜角𝛼等于()A.30°B.60°C.-45°D.135°解析:由参数方程,知𝑥=-2+32𝑡,𝑦=3-32𝑡,两式相加,得直线的普通方程x+y=1,倾斜角为α,则tanα=-1,故α=135°.答案:DZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12【做一做1-2】经过点M(-2,3),倾斜角为3π4的直线𝑙的参数方程是__________.解析:直线l的参数方程为𝑥=-2+𝑡cos3π4,𝑦=3+𝑡sin3π4,即𝑥=-2-22𝑡,𝑦=3+22𝑡.答案:𝑥=-2-22𝑡,𝑦=3+22𝑡ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航122.圆的参数方程若圆心在点M0(x0,y0),半径为R,则圆的参数方程为𝑥=𝑥0+𝑅cos𝜃,𝑦=𝑦0+𝑅sin𝜃(0≤θ≤2π).ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12【做一做2-1】直线3x-4y-9=0与圆𝑥=2cos𝜃,𝑦=2sin𝜃,0≤θ≤2π的位置关系是()A.相切B.相离C.直线过圆心D.相交但直线不过圆心解析:将圆的参数方程化为普通方程为x2+y2=4,所以圆心到直线3x-4y-9=0的距离所以直线与圆相交.又圆心(0,0)不在直线3x-4y-9=0上,故直线与圆相交,但直线不过圆心.答案:Dd=|-9|32+42=952,ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12【做一做2-2】已知圆的方程为x2+y2=4x,则它的参数方程是.解析:x2+y2=4x可化为(x-2)2+y2=4,则圆心为(2,0),半径r=2.故参数方程为𝑥=2+2cos𝜃,𝑦=2sin𝜃(0≤θ≤2π).答案:𝑥=2+2cos𝜃,𝑦=2sin𝜃(0≤θ≤2π)ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航由直线的参数方程求直线的倾斜角剖析如果直线的参数方程是𝑥=𝑥0+𝑡cos𝛼,𝑦=𝑦0+𝑡sin𝛼(𝑡为参数)的形式,由方程直接可得出倾斜角,即方程中的角α,例如,直线的参数方程为𝑥=1+𝑡cos15°,𝑦=1+𝑡sin15°,则直线的倾斜角为15°.如果不是上述形式,例如直线𝑥=1+𝑡sin15°,𝑦=1+𝑡cos15°的倾斜角就不能直接判断了.第一种方法:把参数方程改写为𝑥-1=𝑡sin15°,𝑦-1=𝑡cos15°,消去t,有y-1=1tan15°(𝑥−1),即y-1=tan75°(x-1),故倾斜角为75°.第二种方法:把原方程化为标准形式,即𝑥=1+𝑡cos75°,𝑦=1+𝑡sin75°,可以看出直线的倾斜角为75°.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四题型一参数方程与普通方程互化【例1】把下面直线的参数方程化为普通方程,普通方程化为参数方程.(1)化l1:x+3𝑦−1=0为参数方程;(2)化l2:𝑥=-3+𝑡,𝑦=1+3𝑡为普通方程.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四解:(1)令y=0,得x=1.则直线l1过定点(1,0),k=−13=−33.设直线l1的倾斜角为α,则tanα=−33,𝛼=5π6,cosα=−32,sin𝛼=12.故l1的参数方程为𝑥=1-32𝑡,𝑦=12𝑡.(2)原方程可化为𝑥+3=𝑡,𝑦-1=3𝑡,①②把①代入②得y-1=3(𝑥+3),即l2的普通方程为3𝑥−𝑦+33+1=0.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四反思在(1)(2)中t的几何意义是不同的.在(1)中,|t|=|M0M|(其中点M0的坐标为(1,0),点M(x,y)为直线l1上任意一点).(2)中|t|=12|𝑀0𝑀|(其中点M0的坐标为(-3,1),点M(x,y)为直线l2上任意一点).ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四题型二直线的参数方程与倾斜角【例2】直线𝑥=3+𝑡sin20°,𝑦=-𝑡cos20°的倾斜角是()A.20°B.70°C.110°D.160°ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四解析:方法一:将原方程改写成𝑥-3=𝑡sin20°,-𝑦=𝑡cos20°,消去t,得y=tan110°(x-3),所以直线的倾斜角为110°.方法二:将原参数方程化为𝑥=3+(-𝑡)cos110°,𝑦=(-𝑡)sin110°,令-t=t',则𝑥=3+𝑡'cos110°,𝑦=𝑡'sin110°,所以直线的倾斜角为110°.答案:C反思只有在𝑥=𝑥0+𝑡cos𝛼,𝑦=𝑦0+𝑡sin𝛼(𝑡为参数)中,α才表示直线的倾斜角.若不是这种形式,则需要进行转化.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四题型三直线参数方程的应用【例3】已知直线l:x+y-1=0与抛物线y=x2交于A,B两点,求线段AB的长和点M(-1,2)到A,B两点的距离之积.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四解:因为l过定点M,且l的倾斜角为3π4,所以它的参数方程是𝑥=-1+𝑡cos3π4,𝑦=2+𝑡sin3π4.即𝑥=-1-22𝑡,𝑦=2+22𝑡.①把①代入抛物线方程,得t2+2𝑡−2=0.解得t1=-2+102,𝑡2=-2-102.由参数t的几何意义,得|AB|=|t1-t2|=10,|𝑀𝐴|·|MB|=|t1t2|=2.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四反思本题涉及普通方程和参数方程的互化,在解题过程中,注意参数几何意义的应用.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四题型四圆的参数方程的应用【例4】已知点P(x,y)在圆x2+y2=1上,求x2+2xy+3y2的最大值和最小值.分析利用参数方程,转化成三角函数的问题来解决.解:圆x2+y2=1的参数方程为𝑥=cos𝛼,𝑦=sin𝛼(0≤α≤2π).所以x2+2xy+3y2=cos2α+2cosαsinα+3sin2α=1+cos2𝛼2+sin2α+3×1-cos2𝛼2=2+sin2α-cos2α=2+2sin2𝛼-π4.则当α=3π8或11π8时,x2+2xy+3y2取最大值为2+2;当α=7π8或15π8时,x2+2xy+3y2取最小值为2−2.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型四反思利用参数方程求最值问题是其常见的应用,求解时注意三角公式的应用.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123451.直线𝑥=-2+𝑡cos50°,𝑦=3-𝑡sin40°的倾斜角𝛼等于()A.40°B.50°C.-45°D.135°答案:D解析:根据tanα=-sin40°cos50°=−1,得倾斜角为135°.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123452.若P(2,-1)为圆O:的弦的中点,则该弦所在直线l的方程是()A.x-y-3=0B.x+2y=0C.x+y-1=0D.2x-y-5=0解析:∵圆心O(1,0),∴kPO=-1.∴kl=1.∴直线l方程为x-y-3=0.答案:A𝑥=1+5cos𝜃,𝑦=5sin𝜃(0≤θ≤2π)ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123453.直线𝑥=2-12𝑡,𝑦=-1+12𝑡被圆𝑥2+𝑦2=4截得的弦长为()A.14B.13C.23D.3解析:直线的普通方程为x+y-1=0,圆心到直线的距离d=12=22,所以弦长的一半为22-222=142,故弦长为14.答案:AZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析M