求抛物线y=x^2在点(1,1)的切线方程和法线方程

2025-04-02 10:38:24
推荐回答(5个)
回答1:

曲线的导数就是曲线在点x=xo处的斜率

y=x²,y'=2x

当x=1,y=1,把x值代入y'中

y'(1)=2*1=2

∴切线斜率为2。

用点斜式方程:y-1=2(x-1)

解得切线方程是2x-y-1=0

切线与法线互相垂直,他们乘积为-1,∴法线斜率=-1/2

用点斜式方程:y-1=(-1/2)(x-1)

解得法线方程是x+2y-3=0

简介

P和Q是曲线C上邻近的两点,P是定点,当Q点沿着曲线C无限地接近P点时,割线PQ的极限位置PT叫做曲线C在点P的切线,P点叫做切点;经过切点P并且垂直于切线PT的直线PN叫做曲线C在点P的法线(无限逼近的思想)。

说明:平面几何中,将和圆只有一个公共交点的直线叫做圆的切线.这种定义不适用于一般的曲线;PT是曲线C在点P的切线,但它和曲线C还有另外一个交点;相反,直线l尽管和曲线C只有一个交点,但它却不是曲线C的切线。

回答2:

k=y'=2x=2
切线方程:y-1=2(x-1),即为:y=2x-1

法线方程:y-1=(-1/2)(x-1),即为:y=(3-x)/2

回答3:

曲线的导数就是曲线在点x=xo处的斜率
y=x²,y'=2x
当x=1,y=1,把x值代入y'中
y'(1)=2*1=2
∴切线斜率为2。
用点斜式方程:y-1=2(x-1)
解得切线方程是2x-y-1=0
切线与法线互相垂直,他们乘积为-1,∴法线斜率=-1/2
用点斜式方程:y-1=(-1/2)(x-1)
解得法线方程是x+2y-3=0

回答4:

求导,导函数f'(x)=2x

所以点(1,1)处的切线斜率=f'(1)=2

所以切线方程y=2x+m

代入点(1,1)求m就可以了,最后

y=2x-1

法线与切线垂直,两直线垂直斜率乘积等于-1,

所以法线y=-x/2+n

同样带入(1,1)求n

最后法线y=-x/2+3/2

回答5:

k=y`/x=2=2x/x=2=4
切线方程:y-4=4(x-2)
y=4x-8+4=4x-4
法线方程:k`=-1/k=-1/4
y-4=-1/4(x-2)
4y-16=-x+2
x+4y-18=0

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