Write an equation of each horizontal tangent line to the curve 2y^3+6x^2y-12x^2+6y=1?

Question

I also have to find: the line through the origin with slope-1 is tangent to the curve at point P. Find the x and y coordinates of point P

Answer 1

i'm not going to do it for you but i'll guide you in the right direction.

horizontal tangent lines occur at stationary points. agree?

to find stationary points (turning points) differentiate dy/dx and make ur dy/dx result = 0. When dy/dx = 0 it means there's no slope so we know the curve has to be horizontal there.

u'll probably have something like ay^2+bx+cy+d = 0 and then solve this quadratic and find the roots of it.

after finding ur x roots, sub x values into the curve function to get the corresponding y values. then u have your coordinates.

to get the equation of the tangent , it is simply, y = the y coordinate of the stationary point.



the second part of the question, u already know the gradient of the tangent u need. now get ur dy/dx again and solve for = -1. after u find the x values which have a slop -1, check if their tangents to see if they go through the origin.