Find the equations of the tangent line and the normal line to the curve at the given point.

题目

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)

答案

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