using implicit differentiation,
2(x^2+y^2)(2x+2yy') = 16y'
x(x^2+y^2) + y(x^2+y^2)y' - 4y' = 0
y' = -x(x^2+y^2)/(x^2y + y^3 - 4)
So, the slope is vertical where
x^2y-y^3 -4 = 0
y(x^2-y^2) = 4
Not easy to solve in general, but a little playing around gives
x = ±√3, y=1
(x^2+y^2)^2=16y
Find all values of x where the tangent line is vertical
I found the derivative and looked for where it was undefined but it didn't work
3 answers
Can you show some more steps in between I can't get how to solve for y and x with only 1 equation.
Also isn't there an infinite number of solutions to this equation? How are you supposed to solve it then?