圆锥曲线中的定点、定值问题是高考的热点.笔者最近遇到一些与斜率相关的定点、定值问题,并对一般情形进行研究,可以得到一般性结论,与各位共赏. 定理1:已知点A(x0,y0)是抛物线 y2=2px上的定点,直线l(不过A点)与抛物线交于M、N两点.(1)若 kAM+kAN=c(常数),则直线l斜率为定值;(2)若kAM ·kAN=c(常数),直线l恒过定点. 证明:(1)直线l斜率显然不为0,故设为 x=ty+m,M(x1,y1),N(x2,y2). 由 y2=2px t=ty+m y2-2pty-2pm=0y1+y2=2pt,y1y2=-2pm, kAM +kAN =y1-y0 x1-x0 +y2-y0x2-x0 =2p(y1+y2+2y0) y1y2+y0(y1+y2)+y20 =2p(2pt+2y0) -2pm+2pty0+y20 =c, 即: 4p2t+4py0=-2pmc+2pty0c+y20c, 要斜率为定值,即要 t=-2pmc+y20c-4py0 4p2-2py0c 为定值,所以 c=0,t=-y0p. (i)若A为原点,y0=0,此时直线l斜率不存在; (ii)若A为原点,y0≠0,此时直线l斜率 k=- py0. (2)kAM ·kAN =y1-y0x1-x0 ·y2-y0x2-x0 = 4p2 -2pm+2pty0+y20 =c. 即 -2pmc+2pty0c+ cy20-4p2=0 可解得: m=2pty0c+cy20-4p22mc ,直线l方程为: x=ty+ 2pty0c+cy20-4p22pc, 即2pt(cy+y0c)+cy20-4p2-2pcx=0,恒过定点 (cy20-4p22pc,-y0) . 定理2:已知 A(x0,y0) 是椭圆 x2a2+y2b2=1(a>b>0)上的定点,直线l(不过 A点)与椭圆交于M、N两点.(1)若kAM·kAN=c常数,直线l恒过定点;(2)若 kAM+kAN=c常数,直线l斜率为定值. 证明:(1)若直线l斜率不存在,设 M(t,b1-t2a2),N( t,-b1-t2a2) [HT5,5”]kAM·kAN =(b 1-t2a2 -y0)(-b1- t2a2-y0) (t-x0)2 = [HT] y20-b2(1-t2a2) (t-x0)2 = b2(1- x20a2)-b 2(1-t2a2) (t-x0)2 =b2a2· t+x0t-x0 =c,则b2a2 ·t+x0t-x0 =cx0=0, c=b2a2 才满足. 若直线MN斜率存在,设为 y=kx+m,M(x1,y1),N(x2,y2) b2x2+a2y2-a2b2=0 y=kx+m (a2k2+b2)x2+2a2mkx+ a2m2-a2b2=0 x1+x2= -2a2mka2k2+b2 ,x1x2= a2m2-a2b2 a2k2+b2. kAM ·kAN =y1-y0x1-x0· y2-y0x2-x0 =(kx1+m-y0)(kx2+m-y0) x1x2-x0(x1+x2)+x20= -a2b2k2+m2b2+a2k2y20+b2y20-2my0b2 a2m2-a2b2+2a2mkx0+b2x20+ a2k2x20 = -a2b2k2+m2b2+k2(a2b2-b2x20)+b 2y20-2my0b2 a2m2-a2b2+2a2mkx0+(a2b2-a2y20)+a2k2x20= m2b2-b2x20k2+b2y20-2my0b2 a2m2+2a2mkx0-a2y20+a2k2x20= b2a2 ·m2+y20-2my0-x20k2 m2+2mkx0+k2x20-y20= b2a2· (m-y0+kx0)(m-y0-kx0) (m-y0+kx0)(m+y0+kx0) =b2a2 ·m-y0-kx0 m+y0+kx0 . 所以m= (a2cx0+b2x0)k+a2cy0+b2y0 b2-a2c 直线MN方程为: y=kx+(a2cx0+b2x0)k+a2cy0+b2y0 b2-a2c =k(x+a2cx0+b2x0b2-a2c )+a2cy0+b2y0b2-a2c. 恒过定点 (-a2c+b2 b2-a2c x0,a2c+b2b2-a2c y0). 当定点A(x0,y0)在坐标轴上有: ① 若 x0=0,y0=b直线恒过定点 (0,b(b2+a2c)b2-a2c); ②x0=a,y0=0, 直线恒过定点 (a(a2c+b2) a2c-b2),0) ; ③若x0=0,y0=-b,直线恒过定点 (0,b(b2+a2c)a2c-b2); ④若 x0=-a, y0=0 直线恒过定点 (a(a2c+b2)b2-a2c ,0). (2)若直线MN斜率不存在, kAM+kAN = b1-t2a2-y0t-x0 + -b1-t2a2-y0t-x0 =-2y0t-x0 =c, 则y0=0,c=0才满足. 若直线MN斜率存在, kAM+kAN =y1-y0 x1-x0 +y2-y0x2-x0 = kx1+m-y0x1-x0 +kx2+m-y0x2-x0= [HT5,6]2kx1x2+(m-y0-kx0)(x1+x2)-2mx0+2x0y0 x1x2-x0(x1+x2)+x20= -2a2b2k2-2a2mky0-2b2mx0+2x0y0b2+ 2x0y0a2k2 a2m2-a2b2+2a2mkx0+b2x20+a2k2x20= c. 若直线MN斜率k为定值,则可化为 a2cm2+(2a2ky0-2b2x0-2a2kx0c)m-2a2b2k+2x0y0b2+2x0y0a2k2 +a2b2c-b2x0c-a2k2x20c=0 c=0 2a2ky0-2b2x0-2a2kx0c=0 -2a2b2k+2x0y0b2+2x0y0a2k2=0 ①②③ c=0 k=b2x0a2y0 1,2代入3也成立 当常数c=0时,直线斜率为定值b2x0a2y0. 定理3:已知 A(x0,y0) 是双曲线 x2a2 -y2b2 =1 (a>0,b>0) 上的定点,直线l(不过A点)与椭圆交于M、N两点.(1)若 kAM·kAN =c常数,直线l恒过定点 (a2c-b2 b2+acx0,b2-a2c b2+acy0) ;(2)若 kAM+ kAN=c(=0)常数,直线l斜率为定值 -b2x0a2y0. 证明方法同上. 以上结论的逆命题也是成立的,把对应参数取特值,可以得到一些有用的结论,不再赘述.
圆锥曲线中的定点、定值问题是高考的热点.笔者最近遇到一些与斜率相关的定点、定值问题,并对一般情形进行研究,可以得到一般性结论,与各位共赏. 定理1:已知点A(x0,y0)是抛物线 y2=2px上的定点,直线l(不过A点)与抛物线交于M、N两点.(1)若 kAM+kAN=c(常数),则直线l斜率为定值;(2)若kAM ·kAN=c(常数),直线l恒过定点. 证明:(1)直线l斜率显然不为0,故设为 x=ty+m,M(x1,y1),N(x2,y2). 由 y2=2px t=ty+m y2-2pty-2pm=0y1+y2=2pt,y1y2=-2pm, kAM +kAN =y1-y0 x1-x0 +y2-y0x2-x0 =2p(y1+y2+2y0) y1y2+y0(y1+y2)+y20 =2p(2pt+2y0) -2pm+2pty0+y20 =c, 即: 4p2t+4py0=-2pmc+2pty0c+y20c, 要斜率为定值,即要 t=-2pmc+y20c-4py0 4p2-2py0c 为定值,所以 c=0,t=-y0p. (i)若A为原点,y0=0,此时直线l斜率不存在; (ii)若A为原点,y0≠0,此时直线l斜率 k=- py0. (2)kAM ·kAN =y1-y0x1-x0 ·y2-y0x2-x0 = 4p2 -2pm+2pty0+y20 =c. 即 -2pmc+2pty0c+ cy20-4p2=0 可解得: m=2pty0c+cy20-4p22mc ,直线l方程为: x=ty+ 2pty0c+cy20-4p22pc, 即2pt(cy+y0c)+cy20-4p2-2pcx=0,恒过定点 (cy20-4p22pc,-y0) . 定理2:已知 A(x0,y0) 是椭圆 x2a2+y2b2=1(a>b>0)上的定点,直线l(不过 A点)与椭圆交于M、N两点.(1)若kAM·kAN=c常数,直线l恒过定点;(2)若 kAM+kAN=c常数,直线l斜率为定值. 证明:(1)若直线l斜率不存在,设 M(t,b1-t2a2),N( t,-b1-t2a2) [HT5,5”]kAM·kAN =(b 1-t2a2 -y0)(-b1- t2a2-y0) (t-x0)2 = [HT] y20-b2(1-t2a2) (t-x0)2 = b2(1- x20a2)-b 2(1-t2a2) (t-x0)2 =b2a2· t+x0t-x0 =c,则b2a2 ·t+x0t-x0 =cx0=0, c=b2a2 才满足. 若直线MN斜率存在,设为 y=kx+m,M(x1,y1),N(x2,y2) b2x2+a2y2-a2b2=0 y=kx+m (a2k2+b2)x2+2a2mkx+ a2m2-a2b2=0 x1+x2= -2a2mka2k2+b2 ,x1x2= a2m2-a2b2 a2k2+b2. kAM ·kAN =y1-y0x1-x0· y2-y0x2-x0 =(kx1+m-y0)(kx2+m-y0) x1x2-x0(x1+x2)+x20= -a2b2k2+m2b2+a2k2y20+b2y20-2my0b2 a2m2-a2b2+2a2mkx0+b2x20+ a2k2x20 = -a2b2k2+m2b2+k2(a2b2-b2x20)+b 2y20-2my0b2 a2m2-a2b2+2a2mkx0+(a2b2-a2y20)+a2k2x20= m2b2-b2x20k2+b2y20-2my0b2 a2m2+2a2mkx0-a2y20+a2k2x20= b2a2 ·m2+y20-2my0-x20k2 m2+2mkx0+k2x20-y20= b2a2· (m-y0+kx0)(m-y0-kx0) (m-y0+kx0)(m+y0+kx0) =b2a2 ·m-y0-kx0 m+y0+kx0 . 所以m= (a2cx0+b2x0)k+a2cy0+b2y0 b2-a2c 直线MN方程为: y=kx+(a2cx0+b2x0)k+a2cy0+b2y0 b2-a2c =k(x+a2cx0+b2x0b2-a2c )+a2cy0+b2y0b2-a2c. 恒过定点 (-a2c+b2 b2-a2c x0,a2c+b2b2-a2c y0). 当定点A(x0,y0)在坐标轴上有: ① 若 x0=0,y0=b直线恒过定点 (0,b(b2+a2c)b2-a2c); ②x0=a,y0=0, 直线恒过定点 (a(a2c+b2) a2c-b2),0) ; ③若x0=0,y0=-b,直线恒过定点 (0,b(b2+a2c)a2c-b2); ④若 x0=-a, y0=0 直线恒过定点 (a(a2c+b2)b2-a2c ,0). (2)若直线MN斜率不存在, kAM+kAN = b1-t2a2-y0t-x0 + -b1-t2a2-y0t-x0 =-2y0t-x0 =c, 则y0=0,c=0才满足. 若直线MN斜率存在, kAM+kAN =y1-y0 x1-x0 +y2-y0x2-x0 = kx1+m-y0x1-x0 +kx2+m-y0x2-x0= [HT5,6]2kx1x2+(m-y0-kx0)(x1+x2)-2mx0+2x0y0 x1x2-x0(x1+x2)+x20= -2a2b2k2-2a2mky0-2b2mx0+2x0y0b2+ 2x0y0a2k2 a2m2-a2b2+2a2mkx0+b2x20+a2k2x20= c. 若直线MN斜率k为定值,则可化为 a2cm2+(2a2ky0-2b2x0-2a2kx0c)m-2a2b2k+2x0y0b2+2x0y0a2k2 +a2b2c-b2x0c-a2k2x20c=0 c=0 2a2ky0-2b2x0-2a2kx0c=0 -2a2b2k+2x0y0b2+2x0y0a2k2=0 ①②③ c=0 k=b2x0a2y0 1,2代入3也成立 当常数c=0时,直线斜率为定值b2x0a2y0. 定理3:已知 A(x0,y0) 是双曲线 x2a2 -y2b2 =1 (a>0,b>0) 上的定点,直线l(不过A点)与椭圆交于M、N两点.(1)若 kAM·kAN =c常数,直线l恒过定点 (a2c-b2 b2+acx0,b2-a2c b2+acy0) ;(2)若 kAM+ kAN=c(=0)常数,直线l斜率为定值 -b2x0a2y0. 证明方法同上. 以上结论的逆命题也是成立的,把对应参数取特值,可以得到一些有用的结论,不再赘述.