x^2 - 2xy + y^4 =4
2x - 2y - 2xy' + 4y^3 y' = 0
y' = (x-y)/(x-2y^3)
Plug in (1,-1)
Now find a line through (0,0) whose slope at (h,k) is -1/y' = (h-2k^3)/(k-h)
But such a line has equation y=mx, so we want a line through (h,mh) with slope (h-2m^3h^3)/(mh-h) = (1-2m^3h^2)/(m-1)
Hmmm. I've been having no joy here. Any thoughts?
Given the curve of C defined by the equation x^2 - 2xy +y^4 =4.
1) Find the derivative dy/dx of the curve C by implicit differentiation at the point P= (1,-1).
2) Find a line through the origin that meets the curve perpendicularly.
1 answer
1 answer