求以圆(x-3)²+(y+1)²=8上的点M(5,-3)为切点的切线方程
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![求以圆(x-3)²+(y+1)²=8上的点M(5,-3)为切点的切线方程](/uploads/image/z/3698725-13-5.jpg?t=%E6%B1%82%E4%BB%A5%E5%9C%86%EF%BC%88x-3%EF%BC%89%26%23178%3B%2B%EF%BC%88y%2B1%EF%BC%89%26%23178%3B%3D8%E4%B8%8A%E7%9A%84%E7%82%B9M%EF%BC%885%2C-3%EF%BC%89%E4%B8%BA%E5%88%87%E7%82%B9%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B)
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求以圆(x-3)²+(y+1)²=8上的点M(5,-3)为切点的切线方程
求以圆(x-3)²+(y+1)²=8上的点M(5,-3)为切点的切线方程
求以圆(x-3)²+(y+1)²=8上的点M(5,-3)为切点的切线方程
由圆心O坐标(3,-1)且点M坐标(5,-3)
可得直线OM的斜率为k1=-1
因为OM垂直切线,所以切线斜率:k1乘k2=-1,则k2=1
又过M(5,-3)则又直线的点斜式方程得:y+3=1(x-5)
即:切线方程为x-y-8=0
(x-3)²+(y+1)²=8上的点M(5,-3)
圆心0(3,-1)
KMO=2/-2=-1
K·KMO=-1
∴K=1
∴切线方程:Y+3=x-5
即x-y-8=0
(x-3)²+(y+1)²=8对x求导得:
(x-3)+(y+1)y'=4 ,代入点M(5,-3)
2-2y'=4 y'(5)=-3
点M(5,-3)为切点的切线方程y+3=-(x-5)
圆(x-3)²+(y+1)²=8
过圆心(3,-1),点M(5,-3)的直线方程om斜率k=(-3+1)/(5-3)==-1
切点的切线方程:
y=x+b -3=5+b b=-8
过切点的切线方程:
y=x-8