Find the equations of the tangent line and the normal line to the curve at the given point.
6x^2+3xy+2y^2+17y-6=0,(-1,0)
人气：499 ℃ 时间：2020-02-06 07:43:28
解答
两边求导
12x+3y+3xy'+4yy'+17y'=0
y'=-(12x+3y)/(3x+4y+17)
则过点(-1,0)的切线斜率为
k=-[12*(-1)+3*0]/[3*(-1)+4*0+17]
=12/14
=6/7
所以切线方程为y=(6/7)(x+1)
即6x-7y+6=0
垂线的斜率为k'=-1/k=-7/6
方程为y=-7/6(x+1)
即7x+6y+7=0
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