Question

How do you find the slope of a tangent line to the graph of the function `(x^2+y^2)^2=(x-y)^2` at (1,0) and (1, -1)?

Answer 1

`l_1->y = 1 - x`
`l_2->y= x`

Explanation:

`f(x,y) = (x^2 + y^2)^2 - (x - y)^2=0`

`dy/dx = m(x,y) = -f_x/(f_y) = (x - 2 x^3 - y - 2 x y^2)/(x - y + 2 x^2 y + 2 y^3)`

Given `p_1 = {1,0}` and `p_2 = {1,-1}` we have

`m_1 = m(1,0) = -1`
`m_2=m(1,-1) = 1`

The tangent lines are given by

`l_1->y = 0 + (-1)(x-1) = 1 - x`
`l_2->y = -1 + (1)(x+1) = x`