Find al points where tangent lines to: (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x is either horizontal or vertical using implicit differentiation.

Answer 1

3√(x^2+y^2) * (x + yy') = (x+yy')/√(x^2+y^2) + 1

x+yy'(3√(x^2+y^2) - 1/√(x^2+y^2) = 1

y' = (1/(3√(x^2+y^2) - 1/√(x^2+y^2) - x)/y

y' = [x+√(x^2+y^2) - 3x(x^2+y^2)]/(y(3(x^2+y^2)-1))

So, we want either the numerator or the denominator to be zero.

It might be a bit easier to use polar coordinates.

y' = (x+r-3xr^2)/(3yr^2-y)
= (rcosθ+r-3r^3 cosθ)/(rsinθ(3r^2-1))

r^2 = 1+cosθ

The tangent is vertical where

rsinθ(3r^2-1) = 0
or, at θ = 0 or 2pi

the tangent is horizontal when
r(cosθ+1-3r^2cosθ) = 0
r(cosθ+1-3cosθ(1+cosθ)) = 0
cosθ+1-3cosθ-3cos^2θ = 0
3cos^2θ + 2cosθ - 1 = 0
(3cosθ-1)(cosθ+1) = 0
So, horizontal when
cosθ = -1
cosθ = 1/3

Better double-check the math

Answer 2

Thank you!